Mixing
Solve the sum $sum _{i=1}^4:sum _{j=1}^i:left(icdot :j-1right)$
$begingroup$
Is it correct if this sum is solved this way?
discrete-mathematics proof-verification summation
asked Jan 14 at 4:26
ViktorViktor
477
$endgroup$
add a comment |
$begingroup$
Is it correct if this sum is solved this way?
discrete-mathematics proof-verification summation
asked Jan 14 at 4:26
ViktorViktor
477
$endgroup$
add a comment |
$begingroup$
Is it correct if this sum is solved this way?
discrete-mathematics proof-verification summation
asked Jan 14 at 4:26
ViktorViktor
477
$endgroup$
Is it correct if this sum is solved this way?
discrete-mathematics proof-verification summation
discrete-mathematics proof-verification summation
asked Jan 14 at 4:26
ViktorViktor
477
asked Jan 14 at 4:26
ViktorViktor
477
edited Jan 14 at 4:36
Viktor
asked Jan 14 at 4:26
ViktorViktor
477
asked Jan 14 at 4:26
ViktorViktor
477
asked Jan 14 at 4:26
ViktorViktor
477
477
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Yes, it looks good. You can also check your work on wolfram alpha. Go to wolframalpha.com and enter:
sum i = 1 to 4 sum j = 1 to i (i*j - 1)
as in here.
answered Jan 14 at 5:36
BenBen
3,861616
$endgroup$
add a comment |
$begingroup$
It looks as if you have used the formulas for the sum of a square and a cube, or something. That's great.
But I can never remember those (so I wouldn't be able to tell you if you did it correctly or not). I only remember the Gauß formula, $sum_{i=1}^ni=frac{n(n+1)}2$ (which you did use at one point).
However, since the sum is only up to $4$, you could just do it by hand.
Thus, picking up in the middle, $sum_{i=1}^4frac12(i^3+i^2)-i=frac12(1^3+1^2+2^3+2^2+3^3+3^2+4^3+4^2)-frac{4cdot 5}2=frac12(1+1+8+4+27+9+64+16)-10=frac12(130)-10=65-10=55$.
So, either you lucked out, or you used the formulas correctly.
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
$endgroup$
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
add a comment |
$begingroup$
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=sum_{i=1}^nileft(sum_{j=1}^ij-1right)\
&=sum_{i=1}^nileft(frac {i(i+1)}2-1right)\
&=frac 12sum_{i=1}^ni(i^2+i-2)\
&=frac 12 sum_{i=1}^n i(i-1)(i+2)\
&=frac 12 sum_{i=1}^n(i+1)i(i-1)+i(i-1)\
&=frac 12 sum_{i=1}^n 6binom {i+1}3+2binom i2\
&=3binom {n+2}4+binom {n+1}3\
&=binom{n+1}3left(3cdot frac {n+2}4+1right)\
&=frac 14 binom {n+1}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)end{align}$$
Alternatively,
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=frac 12left(sum_{i=1}^ni^3+sum_{i=1}^ni^2right)-sum_{i=1}^n i\
&=frac 12 left[binom {n+1}2 ^2+frac 13binom {n+1}2 (2n+1)-2binom {n+1}2right]\
&=frac 12binom {n+1}2 left[binom {n+1}2+frac 13 (2n+1)-2right]\
&=frac 12 binom {n+1}2cdot frac16 big[3n(n+1)+2(2n+1)-12big]\
&=frac 1{12} binom {n+1}2cdot (3n^2+7n-10)\
&=frac 1{12}binom {n+1}2 (n-1)(3n+10)\
&=frac 14binom{n+2}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)
end{align}$$
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
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$begingroup$
Yes, it looks good. You can also check your work on wolfram alpha. Go to wolframalpha.com and enter:
sum i = 1 to 4 sum j = 1 to i (i*j - 1)
as in here.
answered Jan 14 at 5:36
BenBen
3,861616
$endgroup$
add a comment |
$begingroup$
Yes, it looks good. You can also check your work on wolfram alpha. Go to wolframalpha.com and enter:
sum i = 1 to 4 sum j = 1 to i (i*j - 1)
as in here.
answered Jan 14 at 5:36
BenBen
3,861616
$endgroup$
add a comment |
$begingroup$
Yes, it looks good. You can also check your work on wolfram alpha. Go to wolframalpha.com and enter:
sum i = 1 to 4 sum j = 1 to i (i*j - 1)
as in here.
answered Jan 14 at 5:36
BenBen
3,861616
$endgroup$
Yes, it looks good. You can also check your work on wolfram alpha. Go to wolframalpha.com and enter:
sum i = 1 to 4 sum j = 1 to i (i*j - 1)
as in here.
answered Jan 14 at 5:36
BenBen
3,861616
answered Jan 14 at 5:36
BenBen
3,861616
answered Jan 14 at 5:36
BenBen
3,861616
answered Jan 14 at 5:36
BenBen
3,861616
3,861616
add a comment |
add a comment |
$begingroup$
It looks as if you have used the formulas for the sum of a square and a cube, or something. That's great.
But I can never remember those (so I wouldn't be able to tell you if you did it correctly or not). I only remember the Gauß formula, $sum_{i=1}^ni=frac{n(n+1)}2$ (which you did use at one point).
However, since the sum is only up to $4$, you could just do it by hand.
Thus, picking up in the middle, $sum_{i=1}^4frac12(i^3+i^2)-i=frac12(1^3+1^2+2^3+2^2+3^3+3^2+4^3+4^2)-frac{4cdot 5}2=frac12(1+1+8+4+27+9+64+16)-10=frac12(130)-10=65-10=55$.
So, either you lucked out, or you used the formulas correctly.
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
$endgroup$
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
add a comment |
$begingroup$
It looks as if you have used the formulas for the sum of a square and a cube, or something. That's great.
But I can never remember those (so I wouldn't be able to tell you if you did it correctly or not). I only remember the Gauß formula, $sum_{i=1}^ni=frac{n(n+1)}2$ (which you did use at one point).
However, since the sum is only up to $4$, you could just do it by hand.
Thus, picking up in the middle, $sum_{i=1}^4frac12(i^3+i^2)-i=frac12(1^3+1^2+2^3+2^2+3^3+3^2+4^3+4^2)-frac{4cdot 5}2=frac12(1+1+8+4+27+9+64+16)-10=frac12(130)-10=65-10=55$.
So, either you lucked out, or you used the formulas correctly.
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
$endgroup$
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
add a comment |
$begingroup$
It looks as if you have used the formulas for the sum of a square and a cube, or something. That's great.
But I can never remember those (so I wouldn't be able to tell you if you did it correctly or not). I only remember the Gauß formula, $sum_{i=1}^ni=frac{n(n+1)}2$ (which you did use at one point).
However, since the sum is only up to $4$, you could just do it by hand.
Thus, picking up in the middle, $sum_{i=1}^4frac12(i^3+i^2)-i=frac12(1^3+1^2+2^3+2^2+3^3+3^2+4^3+4^2)-frac{4cdot 5}2=frac12(1+1+8+4+27+9+64+16)-10=frac12(130)-10=65-10=55$.
So, either you lucked out, or you used the formulas correctly.
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
$endgroup$
It looks as if you have used the formulas for the sum of a square and a cube, or something. That's great.
But I can never remember those (so I wouldn't be able to tell you if you did it correctly or not). I only remember the Gauß formula, $sum_{i=1}^ni=frac{n(n+1)}2$ (which you did use at one point).
However, since the sum is only up to $4$, you could just do it by hand.
Thus, picking up in the middle, $sum_{i=1}^4frac12(i^3+i^2)-i=frac12(1^3+1^2+2^3+2^2+3^3+3^2+4^3+4^2)-frac{4cdot 5}2=frac12(1+1+8+4+27+9+64+16)-10=frac12(130)-10=65-10=55$.
So, either you lucked out, or you used the formulas correctly.
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
answered Jan 14 at 6:38
Chris CusterChris Custer
13.3k3827
13.3k3827
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
add a comment |
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
It's easy to remember them. Only for $i, i^2, i^3$. It's much more easier than going by hand. What if the sum is up to 100 or more.
$endgroup$
– Viktor
Jan 14 at 6:59
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
In that case, yes, we would certainly need them. I would look them up, if it were a big sum.
$endgroup$
– Chris Custer
Jan 14 at 7:00
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
$begingroup$
You're right. They're easy to remember.
$endgroup$
– Chris Custer
Jan 14 at 7:09
add a comment |
$begingroup$
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=sum_{i=1}^nileft(sum_{j=1}^ij-1right)\
&=sum_{i=1}^nileft(frac {i(i+1)}2-1right)\
&=frac 12sum_{i=1}^ni(i^2+i-2)\
&=frac 12 sum_{i=1}^n i(i-1)(i+2)\
&=frac 12 sum_{i=1}^n(i+1)i(i-1)+i(i-1)\
&=frac 12 sum_{i=1}^n 6binom {i+1}3+2binom i2\
&=3binom {n+2}4+binom {n+1}3\
&=binom{n+1}3left(3cdot frac {n+2}4+1right)\
&=frac 14 binom {n+1}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)end{align}$$
Alternatively,
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=frac 12left(sum_{i=1}^ni^3+sum_{i=1}^ni^2right)-sum_{i=1}^n i\
&=frac 12 left[binom {n+1}2 ^2+frac 13binom {n+1}2 (2n+1)-2binom {n+1}2right]\
&=frac 12binom {n+1}2 left[binom {n+1}2+frac 13 (2n+1)-2right]\
&=frac 12 binom {n+1}2cdot frac16 big[3n(n+1)+2(2n+1)-12big]\
&=frac 1{12} binom {n+1}2cdot (3n^2+7n-10)\
&=frac 1{12}binom {n+1}2 (n-1)(3n+10)\
&=frac 14binom{n+2}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)
end{align}$$
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
$endgroup$
add a comment |
$begingroup$
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=sum_{i=1}^nileft(sum_{j=1}^ij-1right)\
&=sum_{i=1}^nileft(frac {i(i+1)}2-1right)\
&=frac 12sum_{i=1}^ni(i^2+i-2)\
&=frac 12 sum_{i=1}^n i(i-1)(i+2)\
&=frac 12 sum_{i=1}^n(i+1)i(i-1)+i(i-1)\
&=frac 12 sum_{i=1}^n 6binom {i+1}3+2binom i2\
&=3binom {n+2}4+binom {n+1}3\
&=binom{n+1}3left(3cdot frac {n+2}4+1right)\
&=frac 14 binom {n+1}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)end{align}$$
Alternatively,
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=frac 12left(sum_{i=1}^ni^3+sum_{i=1}^ni^2right)-sum_{i=1}^n i\
&=frac 12 left[binom {n+1}2 ^2+frac 13binom {n+1}2 (2n+1)-2binom {n+1}2right]\
&=frac 12binom {n+1}2 left[binom {n+1}2+frac 13 (2n+1)-2right]\
&=frac 12 binom {n+1}2cdot frac16 big[3n(n+1)+2(2n+1)-12big]\
&=frac 1{12} binom {n+1}2cdot (3n^2+7n-10)\
&=frac 1{12}binom {n+1}2 (n-1)(3n+10)\
&=frac 14binom{n+2}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)
end{align}$$
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
$endgroup$
add a comment |
$begingroup$
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=sum_{i=1}^nileft(sum_{j=1}^ij-1right)\
&=sum_{i=1}^nileft(frac {i(i+1)}2-1right)\
&=frac 12sum_{i=1}^ni(i^2+i-2)\
&=frac 12 sum_{i=1}^n i(i-1)(i+2)\
&=frac 12 sum_{i=1}^n(i+1)i(i-1)+i(i-1)\
&=frac 12 sum_{i=1}^n 6binom {i+1}3+2binom i2\
&=3binom {n+2}4+binom {n+1}3\
&=binom{n+1}3left(3cdot frac {n+2}4+1right)\
&=frac 14 binom {n+1}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)end{align}$$
Alternatively,
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=frac 12left(sum_{i=1}^ni^3+sum_{i=1}^ni^2right)-sum_{i=1}^n i\
&=frac 12 left[binom {n+1}2 ^2+frac 13binom {n+1}2 (2n+1)-2binom {n+1}2right]\
&=frac 12binom {n+1}2 left[binom {n+1}2+frac 13 (2n+1)-2right]\
&=frac 12 binom {n+1}2cdot frac16 big[3n(n+1)+2(2n+1)-12big]\
&=frac 1{12} binom {n+1}2cdot (3n^2+7n-10)\
&=frac 1{12}binom {n+1}2 (n-1)(3n+10)\
&=frac 14binom{n+2}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)
end{align}$$
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
$endgroup$
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=sum_{i=1}^nileft(sum_{j=1}^ij-1right)\
&=sum_{i=1}^nileft(frac {i(i+1)}2-1right)\
&=frac 12sum_{i=1}^ni(i^2+i-2)\
&=frac 12 sum_{i=1}^n i(i-1)(i+2)\
&=frac 12 sum_{i=1}^n(i+1)i(i-1)+i(i-1)\
&=frac 12 sum_{i=1}^n 6binom {i+1}3+2binom i2\
&=3binom {n+2}4+binom {n+1}3\
&=binom{n+1}3left(3cdot frac {n+2}4+1right)\
&=frac 14 binom {n+1}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)end{align}$$
Alternatively,
$$begin{align}
sum_{i=1}^nsum_{j=1}^i (ij-1)
&=frac 12left(sum_{i=1}^ni^3+sum_{i=1}^ni^2right)-sum_{i=1}^n i\
&=frac 12 left[binom {n+1}2 ^2+frac 13binom {n+1}2 (2n+1)-2binom {n+1}2right]\
&=frac 12binom {n+1}2 left[binom {n+1}2+frac 13 (2n+1)-2right]\
&=frac 12 binom {n+1}2cdot frac16 big[3n(n+1)+2(2n+1)-12big]\
&=frac 1{12} binom {n+1}2cdot (3n^2+7n-10)\
&=frac 1{12}binom {n+1}2 (n-1)(3n+10)\
&=frac 14binom{n+2}3(3n+10)\
&=frac 1{24}(n-1)n(n+1)(3n+10)
end{align}$$
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
answered Jan 14 at 18:26
hypergeometrichypergeometric
17.7k1759
17.7k1759
add a comment |
add a comment |
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