Mixing
Sum of Squares in terms of Sum of Integers
$begingroup$
We know that the sum of squares can be expressed as a multiple of the sum of integers as follows: $$begin{align}
sum_{r=1}^n r^2
&=frac 16 n(n+1)(2n+1)\
&=frac {2n+1}3cdot frac {n(n+1)}2\
&=frac {2n+1}3sum_{r=1}^nrend{align}$$
Is there a simple direct proof to express the sum of squares as $dfrac {2n+1}3$ multiplied by the sum of integers, without first deriving the formula for the sum of squares and then breaking it down as shown above?
summation
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
$begingroup$
We know that the sum of squares can be expressed as a multiple of the sum of integers as follows: $$begin{align}
sum_{r=1}^n r^2
&=frac 16 n(n+1)(2n+1)\
&=frac {2n+1}3cdot frac {n(n+1)}2\
&=frac {2n+1}3sum_{r=1}^nrend{align}$$
Is there a simple direct proof to express the sum of squares as $dfrac {2n+1}3$ multiplied by the sum of integers, without first deriving the formula for the sum of squares and then breaking it down as shown above?
summation
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
$endgroup$
$begingroup$
Possibly not what you are looking for but $$left(sum_{r=1}^{n+1} rright)^2-1 = sum_{r=1}^n (r^3+3r^2+3r+1) = left(sum_{r=1}^n rright)^2 + 3sum_{r=1}^n r^2 + 3sum_{r=1}^n r + n$$ seems a way ..
$endgroup$
– r9m
Aug 3 '15 at 16:20
$begingroup$
That might be a possible lead. Can you develop it further?
$endgroup$
– hypergeometric
Aug 3 '15 at 16:46
$begingroup$
@hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Aug 3 '15 at 16:57
$begingroup$
Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below.
$endgroup$
– hypergeometric
Aug 21 '15 at 6:15
add a comment |
$begingroup$
We know that the sum of squares can be expressed as a multiple of the sum of integers as follows: $$begin{align}
sum_{r=1}^n r^2
&=frac 16 n(n+1)(2n+1)\
&=frac {2n+1}3cdot frac {n(n+1)}2\
&=frac {2n+1}3sum_{r=1}^nrend{align}$$
Is there a simple direct proof to express the sum of squares as $dfrac {2n+1}3$ multiplied by the sum of integers, without first deriving the formula for the sum of squares and then breaking it down as shown above?
summation
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
$endgroup$
We know that the sum of squares can be expressed as a multiple of the sum of integers as follows: $$begin{align}
sum_{r=1}^n r^2
&=frac 16 n(n+1)(2n+1)\
&=frac {2n+1}3cdot frac {n(n+1)}2\
&=frac {2n+1}3sum_{r=1}^nrend{align}$$
Is there a simple direct proof to express the sum of squares as $dfrac {2n+1}3$ multiplied by the sum of integers, without first deriving the formula for the sum of squares and then breaking it down as shown above?
summation
summation
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
edited Aug 3 '15 at 17:25
hypergeometric
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
asked Aug 3 '15 at 16:10
hypergeometrichypergeometric
17.7k1758
17.7k1758
$begingroup$
Possibly not what you are looking for but $$left(sum_{r=1}^{n+1} rright)^2-1 = sum_{r=1}^n (r^3+3r^2+3r+1) = left(sum_{r=1}^n rright)^2 + 3sum_{r=1}^n r^2 + 3sum_{r=1}^n r + n$$ seems a way ..
$endgroup$
– r9m
Aug 3 '15 at 16:20
$begingroup$
That might be a possible lead. Can you develop it further?
$endgroup$
– hypergeometric
Aug 3 '15 at 16:46
$begingroup$
@hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Aug 3 '15 at 16:57
$begingroup$
Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below.
$endgroup$
– hypergeometric
Aug 21 '15 at 6:15
add a comment |
$begingroup$
Possibly not what you are looking for but $$left(sum_{r=1}^{n+1} rright)^2-1 = sum_{r=1}^n (r^3+3r^2+3r+1) = left(sum_{r=1}^n rright)^2 + 3sum_{r=1}^n r^2 + 3sum_{r=1}^n r + n$$ seems a way ..
$endgroup$
– r9m
Aug 3 '15 at 16:20
$begingroup$
That might be a possible lead. Can you develop it further?
$endgroup$
– hypergeometric
Aug 3 '15 at 16:46
$begingroup$
@hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Aug 3 '15 at 16:57
$begingroup$
Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below.
$endgroup$
– hypergeometric
Aug 21 '15 at 6:15
$begingroup$
Possibly not what you are looking for but $$left(sum_{r=1}^{n+1} rright)^2-1 = sum_{r=1}^n (r^3+3r^2+3r+1) = left(sum_{r=1}^n rright)^2 + 3sum_{r=1}^n r^2 + 3sum_{r=1}^n r + n$$ seems a way ..
$endgroup$
– r9m
Aug 3 '15 at 16:20
$begingroup$
Possibly not what you are looking for but $$left(sum_{r=1}^{n+1} rright)^2-1 = sum_{r=1}^n (r^3+3r^2+3r+1) = left(sum_{r=1}^n rright)^2 + 3sum_{r=1}^n r^2 + 3sum_{r=1}^n r + n$$ seems a way ..
$endgroup$
– r9m
Aug 3 '15 at 16:20
$begingroup$
That might be a possible lead. Can you develop it further?
$endgroup$
– hypergeometric
Aug 3 '15 at 16:46
$begingroup$
That might be a possible lead. Can you develop it further?
$endgroup$
– hypergeometric
Aug 3 '15 at 16:46
$begingroup$
@hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Aug 3 '15 at 16:57
$begingroup$
@hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Aug 3 '15 at 16:57
$begingroup$
Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below.
$endgroup$
– hypergeometric
Aug 21 '15 at 6:15
$begingroup$
Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below.
$endgroup$
– hypergeometric
Aug 21 '15 at 6:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$begin{align}
sum_{r=1}^n r^2&=(n+1)^3-1-sum_{r=1}^n (r+1)left((r+1)^2-r^2right)\\
&=(n+1)^3-1-sum_{r=1}^n (r+1)left(2r+1right)\\
&=(n+1)^3-1-2sum_{r=1}^n r^2-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
3sum_{r=1}^n r^2&=(n+1)^3-1-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
sum_{r=1}^n r^2&=frac{n(n+1)}{2}frac{2(n+2)}{3}-sum_{r=1}^{n}r\\
&=frac{n(n+1)}{2}left(frac{2(n+2)}{3}-1right)\\
&=frac{n(n+1)}{2}frac{2n+1}{3}
end{align}$$
ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts. Here, we let $f_k=g_k=k$ and write
$$begin{align}
sum_{k=1}^{n}k^2&=nleft(frac{n(n+1)}{2}right)-sum_{k=0}^{n-1}sum_{ell=0}^{k} k\\
&=nleft(frac{n(n+1)}{2}right)-sum_{k =0}^{n-1}frac{k(k+1)}{2}\\
&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k(k+1)\\
frac32sum_{k=1}^{n}k^2&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k\\
sum_{k=1}^{n}k^2&=frac23 (n+1)left(frac{n(n+1)}{2}right)-frac13sum_{k =1}^n,k\\
&=frac13(2n+1)left(frac{n(n+1)}{2}right)
end{align}$$
again as expected!
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
$endgroup$
– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
$endgroup$
– Mark Viola
Aug 3 '15 at 17:50
$begingroup$
Please leave the solution here. It helps to have different approaches.
$endgroup$
– hypergeometric
Aug 4 '15 at 2:59
$begingroup$
@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
$endgroup$
– Mark Viola
Aug 4 '15 at 3:30
add a comment |
$begingroup$
New Solution
Have just found a proof which does not require the prior knowledge of the result of the sum of integers.
$$begin{align}
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i(2j-1)&& ...(1)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(n-i)+1&& ...(2)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(i-j)+1&& ...(3)\
(1)+(2)+(3):\
3sum_{i=1}^n i^2
&=sum_{i=1}^nsum_{j=1}^i (2j-1)+2(n-1)+1+2(i-j)+1\
&=sum_{i=1}^nsum_{j=1}^i (2n+1)\
&=(2n+1)sum_{i=1}^n i\
sum_{i=1}^n i^2&=frac{2n+1}3sum_{i=1}^n iqquadblacksquare
end{align}$$
This proof is a transcription of the diagrammatic proof of the same as shown on the wikipedia page here.
See also the nice diagrams in this solution here.
Earlier post shown below
This is too long for a comment so it's being posted in the solution section.
A synthetic and rather cumbersome approach might be as follows:
$$begin{align}
(2m+1)sum_{r=1}^mr-(2m-1)sum_{r=1}^{m-1}r
&=(2m+1)frac {m(m+1)}2-(2m-1)frac{m(m-1)}2\
&=frac m2left[(2m+1)(m+1)-(2m-1)(m-1)right]\
&=3m^2end{align}$$
Summing $m$ from $1$ to $n$ and telescoping LHS gives
$$(2n+1)sum_{r=1}^nr=3sum_{m=1}^nm^2color{lightgray}{=3sum_{r=1}^n r^2}\
sum_{r=1}^nr^2=frac {2n+1}3sum_{r=1}^nrqquadblacksquare$$
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$begin{align}
sum_{r=1}^n r^2&=(n+1)^3-1-sum_{r=1}^n (r+1)left((r+1)^2-r^2right)\\
&=(n+1)^3-1-sum_{r=1}^n (r+1)left(2r+1right)\\
&=(n+1)^3-1-2sum_{r=1}^n r^2-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
3sum_{r=1}^n r^2&=(n+1)^3-1-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
sum_{r=1}^n r^2&=frac{n(n+1)}{2}frac{2(n+2)}{3}-sum_{r=1}^{n}r\\
&=frac{n(n+1)}{2}left(frac{2(n+2)}{3}-1right)\\
&=frac{n(n+1)}{2}frac{2n+1}{3}
end{align}$$
ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts. Here, we let $f_k=g_k=k$ and write
$$begin{align}
sum_{k=1}^{n}k^2&=nleft(frac{n(n+1)}{2}right)-sum_{k=0}^{n-1}sum_{ell=0}^{k} k\\
&=nleft(frac{n(n+1)}{2}right)-sum_{k =0}^{n-1}frac{k(k+1)}{2}\\
&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k(k+1)\\
frac32sum_{k=1}^{n}k^2&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k\\
sum_{k=1}^{n}k^2&=frac23 (n+1)left(frac{n(n+1)}{2}right)-frac13sum_{k =1}^n,k\\
&=frac13(2n+1)left(frac{n(n+1)}{2}right)
end{align}$$
again as expected!
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
$endgroup$
– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
$endgroup$
– Mark Viola
Aug 3 '15 at 17:50
$begingroup$
Please leave the solution here. It helps to have different approaches.
$endgroup$
– hypergeometric
Aug 4 '15 at 2:59
$begingroup$
@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
$endgroup$
– Mark Viola
Aug 4 '15 at 3:30
add a comment |
$begingroup$
Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$begin{align}
sum_{r=1}^n r^2&=(n+1)^3-1-sum_{r=1}^n (r+1)left((r+1)^2-r^2right)\\
&=(n+1)^3-1-sum_{r=1}^n (r+1)left(2r+1right)\\
&=(n+1)^3-1-2sum_{r=1}^n r^2-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
3sum_{r=1}^n r^2&=(n+1)^3-1-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
sum_{r=1}^n r^2&=frac{n(n+1)}{2}frac{2(n+2)}{3}-sum_{r=1}^{n}r\\
&=frac{n(n+1)}{2}left(frac{2(n+2)}{3}-1right)\\
&=frac{n(n+1)}{2}frac{2n+1}{3}
end{align}$$
ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts. Here, we let $f_k=g_k=k$ and write
$$begin{align}
sum_{k=1}^{n}k^2&=nleft(frac{n(n+1)}{2}right)-sum_{k=0}^{n-1}sum_{ell=0}^{k} k\\
&=nleft(frac{n(n+1)}{2}right)-sum_{k =0}^{n-1}frac{k(k+1)}{2}\\
&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k(k+1)\\
frac32sum_{k=1}^{n}k^2&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k\\
sum_{k=1}^{n}k^2&=frac23 (n+1)left(frac{n(n+1)}{2}right)-frac13sum_{k =1}^n,k\\
&=frac13(2n+1)left(frac{n(n+1)}{2}right)
end{align}$$
again as expected!
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
$endgroup$
– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
$endgroup$
– Mark Viola
Aug 3 '15 at 17:50
$begingroup$
Please leave the solution here. It helps to have different approaches.
$endgroup$
– hypergeometric
Aug 4 '15 at 2:59
$begingroup$
@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
$endgroup$
– Mark Viola
Aug 4 '15 at 3:30
add a comment |
$begingroup$
Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$begin{align}
sum_{r=1}^n r^2&=(n+1)^3-1-sum_{r=1}^n (r+1)left((r+1)^2-r^2right)\\
&=(n+1)^3-1-sum_{r=1}^n (r+1)left(2r+1right)\\
&=(n+1)^3-1-2sum_{r=1}^n r^2-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
3sum_{r=1}^n r^2&=(n+1)^3-1-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
sum_{r=1}^n r^2&=frac{n(n+1)}{2}frac{2(n+2)}{3}-sum_{r=1}^{n}r\\
&=frac{n(n+1)}{2}left(frac{2(n+2)}{3}-1right)\\
&=frac{n(n+1)}{2}frac{2n+1}{3}
end{align}$$
ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts. Here, we let $f_k=g_k=k$ and write
$$begin{align}
sum_{k=1}^{n}k^2&=nleft(frac{n(n+1)}{2}right)-sum_{k=0}^{n-1}sum_{ell=0}^{k} k\\
&=nleft(frac{n(n+1)}{2}right)-sum_{k =0}^{n-1}frac{k(k+1)}{2}\\
&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k(k+1)\\
frac32sum_{k=1}^{n}k^2&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k\\
sum_{k=1}^{n}k^2&=frac23 (n+1)left(frac{n(n+1)}{2}right)-frac13sum_{k =1}^n,k\\
&=frac13(2n+1)left(frac{n(n+1)}{2}right)
end{align}$$
again as expected!
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
$endgroup$
Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$begin{align}
sum_{r=1}^n r^2&=(n+1)^3-1-sum_{r=1}^n (r+1)left((r+1)^2-r^2right)\\
&=(n+1)^3-1-sum_{r=1}^n (r+1)left(2r+1right)\\
&=(n+1)^3-1-2sum_{r=1}^n r^2-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
3sum_{r=1}^n r^2&=(n+1)^3-1-3sum_{r=1}^{n}r-sum_{r=1}^{n}1\\
sum_{r=1}^n r^2&=frac{n(n+1)}{2}frac{2(n+2)}{3}-sum_{r=1}^{n}r\\
&=frac{n(n+1)}{2}left(frac{2(n+2)}{3}-1right)\\
&=frac{n(n+1)}{2}frac{2n+1}{3}
end{align}$$
ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts. Here, we let $f_k=g_k=k$ and write
$$begin{align}
sum_{k=1}^{n}k^2&=nleft(frac{n(n+1)}{2}right)-sum_{k=0}^{n-1}sum_{ell=0}^{k} k\\
&=nleft(frac{n(n+1)}{2}right)-sum_{k =0}^{n-1}frac{k(k+1)}{2}\\
&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k(k+1)\\
frac32sum_{k=1}^{n}k^2&=(n+1)left(frac{n(n+1)}{2}right)-frac12sum_{k =1}^n k\\
sum_{k=1}^{n}k^2&=frac23 (n+1)left(frac{n(n+1)}{2}right)-frac13sum_{k =1}^n,k\\
&=frac13(2n+1)left(frac{n(n+1)}{2}right)
end{align}$$
again as expected!
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
edited Aug 4 '15 at 3:24
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
answered Aug 3 '15 at 16:56
Mark ViolaMark Viola
132k1275173
132k1275173
$begingroup$
Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
$endgroup$
– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
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– Mark Viola
Aug 3 '15 at 17:50
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Please leave the solution here. It helps to have different approaches.
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– hypergeometric
Aug 4 '15 at 2:59
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@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
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– Mark Viola
Aug 4 '15 at 3:30
add a comment |
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Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
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– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
$endgroup$
– Mark Viola
Aug 3 '15 at 17:50
$begingroup$
Please leave the solution here. It helps to have different approaches.
$endgroup$
– hypergeometric
Aug 4 '15 at 2:59
$begingroup$
@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
$endgroup$
– Mark Viola
Aug 4 '15 at 3:30
$begingroup$
Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
$endgroup$
– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3sum r^2+3sum r+n$$.
$endgroup$
– hypergeometric
Aug 3 '15 at 17:21
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
$endgroup$
– Mark Viola
Aug 3 '15 at 17:50
$begingroup$
@hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help.
$endgroup$
– Mark Viola
Aug 3 '15 at 17:50
$begingroup$
Please leave the solution here. It helps to have different approaches.
$endgroup$
– hypergeometric
Aug 4 '15 at 2:59
$begingroup$
Please leave the solution here. It helps to have different approaches.
$endgroup$
– hypergeometric
Aug 4 '15 at 2:59
$begingroup$
@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
$endgroup$
– Mark Viola
Aug 4 '15 at 3:30
$begingroup$
@hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking.
$endgroup$
– Mark Viola
Aug 4 '15 at 3:30
add a comment |
$begingroup$
New Solution
Have just found a proof which does not require the prior knowledge of the result of the sum of integers.
$$begin{align}
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i(2j-1)&& ...(1)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(n-i)+1&& ...(2)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(i-j)+1&& ...(3)\
(1)+(2)+(3):\
3sum_{i=1}^n i^2
&=sum_{i=1}^nsum_{j=1}^i (2j-1)+2(n-1)+1+2(i-j)+1\
&=sum_{i=1}^nsum_{j=1}^i (2n+1)\
&=(2n+1)sum_{i=1}^n i\
sum_{i=1}^n i^2&=frac{2n+1}3sum_{i=1}^n iqquadblacksquare
end{align}$$
This proof is a transcription of the diagrammatic proof of the same as shown on the wikipedia page here.
See also the nice diagrams in this solution here.
Earlier post shown below
This is too long for a comment so it's being posted in the solution section.
A synthetic and rather cumbersome approach might be as follows:
$$begin{align}
(2m+1)sum_{r=1}^mr-(2m-1)sum_{r=1}^{m-1}r
&=(2m+1)frac {m(m+1)}2-(2m-1)frac{m(m-1)}2\
&=frac m2left[(2m+1)(m+1)-(2m-1)(m-1)right]\
&=3m^2end{align}$$
Summing $m$ from $1$ to $n$ and telescoping LHS gives
$$(2n+1)sum_{r=1}^nr=3sum_{m=1}^nm^2color{lightgray}{=3sum_{r=1}^n r^2}\
sum_{r=1}^nr^2=frac {2n+1}3sum_{r=1}^nrqquadblacksquare$$
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
$begingroup$
New Solution
Have just found a proof which does not require the prior knowledge of the result of the sum of integers.
$$begin{align}
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i(2j-1)&& ...(1)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(n-i)+1&& ...(2)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(i-j)+1&& ...(3)\
(1)+(2)+(3):\
3sum_{i=1}^n i^2
&=sum_{i=1}^nsum_{j=1}^i (2j-1)+2(n-1)+1+2(i-j)+1\
&=sum_{i=1}^nsum_{j=1}^i (2n+1)\
&=(2n+1)sum_{i=1}^n i\
sum_{i=1}^n i^2&=frac{2n+1}3sum_{i=1}^n iqquadblacksquare
end{align}$$
This proof is a transcription of the diagrammatic proof of the same as shown on the wikipedia page here.
See also the nice diagrams in this solution here.
Earlier post shown below
This is too long for a comment so it's being posted in the solution section.
A synthetic and rather cumbersome approach might be as follows:
$$begin{align}
(2m+1)sum_{r=1}^mr-(2m-1)sum_{r=1}^{m-1}r
&=(2m+1)frac {m(m+1)}2-(2m-1)frac{m(m-1)}2\
&=frac m2left[(2m+1)(m+1)-(2m-1)(m-1)right]\
&=3m^2end{align}$$
Summing $m$ from $1$ to $n$ and telescoping LHS gives
$$(2n+1)sum_{r=1}^nr=3sum_{m=1}^nm^2color{lightgray}{=3sum_{r=1}^n r^2}\
sum_{r=1}^nr^2=frac {2n+1}3sum_{r=1}^nrqquadblacksquare$$
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
$begingroup$
New Solution
Have just found a proof which does not require the prior knowledge of the result of the sum of integers.
$$begin{align}
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i(2j-1)&& ...(1)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(n-i)+1&& ...(2)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(i-j)+1&& ...(3)\
(1)+(2)+(3):\
3sum_{i=1}^n i^2
&=sum_{i=1}^nsum_{j=1}^i (2j-1)+2(n-1)+1+2(i-j)+1\
&=sum_{i=1}^nsum_{j=1}^i (2n+1)\
&=(2n+1)sum_{i=1}^n i\
sum_{i=1}^n i^2&=frac{2n+1}3sum_{i=1}^n iqquadblacksquare
end{align}$$
This proof is a transcription of the diagrammatic proof of the same as shown on the wikipedia page here.
See also the nice diagrams in this solution here.
Earlier post shown below
This is too long for a comment so it's being posted in the solution section.
A synthetic and rather cumbersome approach might be as follows:
$$begin{align}
(2m+1)sum_{r=1}^mr-(2m-1)sum_{r=1}^{m-1}r
&=(2m+1)frac {m(m+1)}2-(2m-1)frac{m(m-1)}2\
&=frac m2left[(2m+1)(m+1)-(2m-1)(m-1)right]\
&=3m^2end{align}$$
Summing $m$ from $1$ to $n$ and telescoping LHS gives
$$(2n+1)sum_{r=1}^nr=3sum_{m=1}^nm^2color{lightgray}{=3sum_{r=1}^n r^2}\
sum_{r=1}^nr^2=frac {2n+1}3sum_{r=1}^nrqquadblacksquare$$
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
$endgroup$
New Solution
Have just found a proof which does not require the prior knowledge of the result of the sum of integers.
$$begin{align}
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i(2j-1)&& ...(1)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(n-i)+1&& ...(2)\
sum_{i=1}^n i^2&=sum_{i=1}^nsum_{j=1}^i 2(i-j)+1&& ...(3)\
(1)+(2)+(3):\
3sum_{i=1}^n i^2
&=sum_{i=1}^nsum_{j=1}^i (2j-1)+2(n-1)+1+2(i-j)+1\
&=sum_{i=1}^nsum_{j=1}^i (2n+1)\
&=(2n+1)sum_{i=1}^n i\
sum_{i=1}^n i^2&=frac{2n+1}3sum_{i=1}^n iqquadblacksquare
end{align}$$
This proof is a transcription of the diagrammatic proof of the same as shown on the wikipedia page here.
See also the nice diagrams in this solution here.
Earlier post shown below
This is too long for a comment so it's being posted in the solution section.
A synthetic and rather cumbersome approach might be as follows:
$$begin{align}
(2m+1)sum_{r=1}^mr-(2m-1)sum_{r=1}^{m-1}r
&=(2m+1)frac {m(m+1)}2-(2m-1)frac{m(m-1)}2\
&=frac m2left[(2m+1)(m+1)-(2m-1)(m-1)right]\
&=3m^2end{align}$$
Summing $m$ from $1$ to $n$ and telescoping LHS gives
$$(2n+1)sum_{r=1}^nr=3sum_{m=1}^nm^2color{lightgray}{=3sum_{r=1}^n r^2}\
sum_{r=1}^nr^2=frac {2n+1}3sum_{r=1}^nrqquadblacksquare$$
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
edited Jan 11 at 21:19
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
answered Aug 21 '15 at 16:34
hypergeometrichypergeometric
17.7k1758
17.7k1758
add a comment |
add a comment |
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$begingroup$
Possibly not what you are looking for but $$left(sum_{r=1}^{n+1} rright)^2-1 = sum_{r=1}^n (r^3+3r^2+3r+1) = left(sum_{r=1}^n rright)^2 + 3sum_{r=1}^n r^2 + 3sum_{r=1}^n r + n$$ seems a way ..
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– r9m
Aug 3 '15 at 16:20
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That might be a possible lead. Can you develop it further?
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– hypergeometric
Aug 3 '15 at 16:46
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@hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Aug 3 '15 at 16:57
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Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below.
$endgroup$
– hypergeometric
Aug 21 '15 at 6:15