Mixing
Solving a summation using summation rules
$begingroup$
I am given two problems.
$$sum_{i=0}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 1$$
$$ sum_{i=1}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 2$$
I am asked to solve it.
I know I can manipulate this into three separate summations:
Attempt:
$$ sum_{i=0}^{n} (4i) + sum_{i=0}^{n} (frac{3}{4}n) + sum_{i=0}^{n} (frac{1}{2}) tag 1 $$
Unfortunately, I am confused on how to proceed other than the fact that the third summation will turn into $ frac{1}{2}$. (because there is no i or n)
May anyone share with me the next step? I'm not sure in what form the solution should be in.
EDIT:
would the first term turn into: $ 4 times frac{n(n+1)}{2}$
EDIT 2:
I've currently got
$$ left(4 times frac{n(n+1)}{2}right) + (?) + frac{1}{2}n tag 2$$
summation
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
$endgroup$
add a comment |
$begingroup$
I am given two problems.
$$sum_{i=0}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 1$$
$$ sum_{i=1}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 2$$
I am asked to solve it.
I know I can manipulate this into three separate summations:
Attempt:
$$ sum_{i=0}^{n} (4i) + sum_{i=0}^{n} (frac{3}{4}n) + sum_{i=0}^{n} (frac{1}{2}) tag 1 $$
Unfortunately, I am confused on how to proceed other than the fact that the third summation will turn into $ frac{1}{2}$. (because there is no i or n)
May anyone share with me the next step? I'm not sure in what form the solution should be in.
EDIT:
would the first term turn into: $ 4 times frac{n(n+1)}{2}$
EDIT 2:
I've currently got
$$ left(4 times frac{n(n+1)}{2}right) + (?) + frac{1}{2}n tag 2$$
summation
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
$endgroup$
$begingroup$
@nickD the starting point of the lower limit is different
$endgroup$
– Arthur Green
Jan 24 at 18:57
1
$begingroup$
Yes to the last prompt. The second and third terms are independent of $i$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Jan 24 at 18:57
1
$begingroup$
Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $frac{1}{2}$: it is a sum of $n$ of them.
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
... or $n+1$ of them, depending on the lower limit (which I missed - sorry).
$endgroup$
– NickD
Jan 24 at 18:59
2
$begingroup$
Hint: $sum_{i=1}^n 1 = n$.
$endgroup$
– NickD
Jan 24 at 19:01
add a comment |
$begingroup$
I am given two problems.
$$sum_{i=0}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 1$$
$$ sum_{i=1}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 2$$
I am asked to solve it.
I know I can manipulate this into three separate summations:
Attempt:
$$ sum_{i=0}^{n} (4i) + sum_{i=0}^{n} (frac{3}{4}n) + sum_{i=0}^{n} (frac{1}{2}) tag 1 $$
Unfortunately, I am confused on how to proceed other than the fact that the third summation will turn into $ frac{1}{2}$. (because there is no i or n)
May anyone share with me the next step? I'm not sure in what form the solution should be in.
EDIT:
would the first term turn into: $ 4 times frac{n(n+1)}{2}$
EDIT 2:
I've currently got
$$ left(4 times frac{n(n+1)}{2}right) + (?) + frac{1}{2}n tag 2$$
summation
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
$endgroup$
I am given two problems.
$$sum_{i=0}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 1$$
$$ sum_{i=1}^{n} (4i + frac{3}{4}n + frac{1}{2}) tag 2$$
I am asked to solve it.
I know I can manipulate this into three separate summations:
Attempt:
$$ sum_{i=0}^{n} (4i) + sum_{i=0}^{n} (frac{3}{4}n) + sum_{i=0}^{n} (frac{1}{2}) tag 1 $$
Unfortunately, I am confused on how to proceed other than the fact that the third summation will turn into $ frac{1}{2}$. (because there is no i or n)
May anyone share with me the next step? I'm not sure in what form the solution should be in.
EDIT:
would the first term turn into: $ 4 times frac{n(n+1)}{2}$
EDIT 2:
I've currently got
$$ left(4 times frac{n(n+1)}{2}right) + (?) + frac{1}{2}n tag 2$$
summation
summation
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
edited Jan 24 at 21:49
Viktor Glombik
1,1771528
1,1771528
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
asked Jan 24 at 18:53
Arthur GreenArthur Green
796
796
$begingroup$
@nickD the starting point of the lower limit is different
$endgroup$
– Arthur Green
Jan 24 at 18:57
1
$begingroup$
Yes to the last prompt. The second and third terms are independent of $i$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Jan 24 at 18:57
1
$begingroup$
Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $frac{1}{2}$: it is a sum of $n$ of them.
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
... or $n+1$ of them, depending on the lower limit (which I missed - sorry).
$endgroup$
– NickD
Jan 24 at 18:59
2
$begingroup$
Hint: $sum_{i=1}^n 1 = n$.
$endgroup$
– NickD
Jan 24 at 19:01
add a comment |
$begingroup$
@nickD the starting point of the lower limit is different
$endgroup$
– Arthur Green
Jan 24 at 18:57
1
$begingroup$
Yes to the last prompt. The second and third terms are independent of $i$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Jan 24 at 18:57
1
$begingroup$
Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $frac{1}{2}$: it is a sum of $n$ of them.
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
... or $n+1$ of them, depending on the lower limit (which I missed - sorry).
$endgroup$
– NickD
Jan 24 at 18:59
2
$begingroup$
Hint: $sum_{i=1}^n 1 = n$.
$endgroup$
– NickD
Jan 24 at 19:01
$begingroup$
@nickD the starting point of the lower limit is different
$endgroup$
– Arthur Green
Jan 24 at 18:57
$begingroup$
@nickD the starting point of the lower limit is different
$endgroup$
– Arthur Green
Jan 24 at 18:57
1
1
$begingroup$
Yes to the last prompt. The second and third terms are independent of $i$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Jan 24 at 18:57
$begingroup$
Yes to the last prompt. The second and third terms are independent of $i$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Jan 24 at 18:57
1
1
$begingroup$
Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $frac{1}{2}$: it is a sum of $n$ of them.
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $frac{1}{2}$: it is a sum of $n$ of them.
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
... or $n+1$ of them, depending on the lower limit (which I missed - sorry).
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
... or $n+1$ of them, depending on the lower limit (which I missed - sorry).
$endgroup$
– NickD
Jan 24 at 18:59
2
2
$begingroup$
Hint: $sum_{i=1}^n 1 = n$.
$endgroup$
– NickD
Jan 24 at 19:01
$begingroup$
Hint: $sum_{i=1}^n 1 = n$.
$endgroup$
– NickD
Jan 24 at 19:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the OP has parenthesized things correctly in the question, the answer by Viktor Glombik is WRONG. It should go like this:
begin{align*}
sum_{i=0}^{n} left(4i + frac{3}{4}n + frac{1}{2}right)
& = sum_{i=0}^{n} left(4i right) + sum_{i=0}^{n}frac{3}{4}n + sum_{i=0}^{n}frac{1}{2} \
&= 4 sum_{i=0}^{n} i + frac{3}{4}nsum_{i=0}^{n} 1 + frac{1}{2}sum_{i=0}^{n} 1\
& = 4 frac{n(n+1)}{2} + frac{3}{4}n(n+1)+ frac{1}{2}(n+1) \
& = 2n (n + 1) + frac{3}{4}n(n+1)+ frac{1}{2}(n+1)\
& = frac{11}{4} n(n+1) + frac{1}{2} (n+1)
end{align*}
answered Jan 24 at 20:09
NickDNickD
1,1921512
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If the OP has parenthesized things correctly in the question, the answer by Viktor Glombik is WRONG. It should go like this:
begin{align*}
sum_{i=0}^{n} left(4i + frac{3}{4}n + frac{1}{2}right)
& = sum_{i=0}^{n} left(4i right) + sum_{i=0}^{n}frac{3}{4}n + sum_{i=0}^{n}frac{1}{2} \
&= 4 sum_{i=0}^{n} i + frac{3}{4}nsum_{i=0}^{n} 1 + frac{1}{2}sum_{i=0}^{n} 1\
& = 4 frac{n(n+1)}{2} + frac{3}{4}n(n+1)+ frac{1}{2}(n+1) \
& = 2n (n + 1) + frac{3}{4}n(n+1)+ frac{1}{2}(n+1)\
& = frac{11}{4} n(n+1) + frac{1}{2} (n+1)
end{align*}
answered Jan 24 at 20:09
NickDNickD
1,1921512
$endgroup$
add a comment |
$begingroup$
If the OP has parenthesized things correctly in the question, the answer by Viktor Glombik is WRONG. It should go like this:
begin{align*}
sum_{i=0}^{n} left(4i + frac{3}{4}n + frac{1}{2}right)
& = sum_{i=0}^{n} left(4i right) + sum_{i=0}^{n}frac{3}{4}n + sum_{i=0}^{n}frac{1}{2} \
&= 4 sum_{i=0}^{n} i + frac{3}{4}nsum_{i=0}^{n} 1 + frac{1}{2}sum_{i=0}^{n} 1\
& = 4 frac{n(n+1)}{2} + frac{3}{4}n(n+1)+ frac{1}{2}(n+1) \
& = 2n (n + 1) + frac{3}{4}n(n+1)+ frac{1}{2}(n+1)\
& = frac{11}{4} n(n+1) + frac{1}{2} (n+1)
end{align*}
answered Jan 24 at 20:09
NickDNickD
1,1921512
$endgroup$
add a comment |
$begingroup$
If the OP has parenthesized things correctly in the question, the answer by Viktor Glombik is WRONG. It should go like this:
begin{align*}
sum_{i=0}^{n} left(4i + frac{3}{4}n + frac{1}{2}right)
& = sum_{i=0}^{n} left(4i right) + sum_{i=0}^{n}frac{3}{4}n + sum_{i=0}^{n}frac{1}{2} \
&= 4 sum_{i=0}^{n} i + frac{3}{4}nsum_{i=0}^{n} 1 + frac{1}{2}sum_{i=0}^{n} 1\
& = 4 frac{n(n+1)}{2} + frac{3}{4}n(n+1)+ frac{1}{2}(n+1) \
& = 2n (n + 1) + frac{3}{4}n(n+1)+ frac{1}{2}(n+1)\
& = frac{11}{4} n(n+1) + frac{1}{2} (n+1)
end{align*}
answered Jan 24 at 20:09
NickDNickD
1,1921512
$endgroup$
If the OP has parenthesized things correctly in the question, the answer by Viktor Glombik is WRONG. It should go like this:
begin{align*}
sum_{i=0}^{n} left(4i + frac{3}{4}n + frac{1}{2}right)
& = sum_{i=0}^{n} left(4i right) + sum_{i=0}^{n}frac{3}{4}n + sum_{i=0}^{n}frac{1}{2} \
&= 4 sum_{i=0}^{n} i + frac{3}{4}nsum_{i=0}^{n} 1 + frac{1}{2}sum_{i=0}^{n} 1\
& = 4 frac{n(n+1)}{2} + frac{3}{4}n(n+1)+ frac{1}{2}(n+1) \
& = 2n (n + 1) + frac{3}{4}n(n+1)+ frac{1}{2}(n+1)\
& = frac{11}{4} n(n+1) + frac{1}{2} (n+1)
end{align*}
answered Jan 24 at 20:09
NickDNickD
1,1921512
answered Jan 24 at 20:09
NickDNickD
1,1921512
answered Jan 24 at 20:09
NickDNickD
1,1921512
answered Jan 24 at 20:09
NickDNickD
1,1921512
1,1921512
add a comment |
add a comment |
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$begingroup$
@nickD the starting point of the lower limit is different
$endgroup$
– Arthur Green
Jan 24 at 18:57
1
$begingroup$
Yes to the last prompt. The second and third terms are independent of $i$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Jan 24 at 18:57
1
$begingroup$
Note that the summation is over $i$ and the summands of the second and third summation do not depend on $i$. N.B. The third sum will NOT turn into $frac{1}{2}$: it is a sum of $n$ of them.
$endgroup$
– NickD
Jan 24 at 18:59
$begingroup$
... or $n+1$ of them, depending on the lower limit (which I missed - sorry).
$endgroup$
– NickD
Jan 24 at 18:59
2
$begingroup$
Hint: $sum_{i=1}^n 1 = n$.
$endgroup$
– NickD
Jan 24 at 19:01