Sum of series of upto $(2n+1)$ terms
The simplest method would be to split this sequence into two different sequences. The first is each odd term and it gives: $ 1^2 + 2^2 + 3^2 + 4^2 + ...$. The second is every odd term and this gives: $ 1^2.2 + 2^2.3 + 3^2.4 + 4^2.5 + ...$. The $(2n+1)^{th}$ term is an odd term, so it is from the first sequence. So, the first sequence goes up to $(n+1)$ terms and the second goes up to $n$ terms. Check this: $(n+1)$ + $n$ = $2n+1$.
The sum to $n$ terms of the first sequence is given by: $\sum_{k=0}^{n} n^2 $ which is equal to $\frac{n}{6}(n+1)(2n+1)$. so, the sum of $(n+1)$ terms is $\frac{n}{6}(n+1)(2n+1) + (n+1)^2 $.
The sum of the second sequence is given by: $\sum_{k=0}^{n} n^2 (n+1) $ = $\sum_{k=0}^{n} (n^3 + n^2) $ = $\sum_{k=0}^{n} n^3 $ + $\sum_{k=0}^{n} n^2$. We already know the value of the second part(as we did it above). The first part is $\sum_{k=0}^{n} n^3 $ = $ \frac{n^2(n+1)^2}{4}$.
So, the total sum is given by: $\frac{n}{6}(n+1)(2n+1) + (n+1)^2 $ + $\frac{n}{6}(n+1)(2n+1)$ + $ \frac{n^2(n+1)^2}{4}$. I will leave you to simplify the final answer.
http://pirate.shu.edu/~wachsmut/ira/infinity/answers/sm_sq_cb.html
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Way to infinity
Updated on August 29, 2020Comments
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Way to infinity about 3 years
Recently I have found the following series . I have to determine the sum of up to ($2n+1$) term of the following series:
$1+2(1^2)+2^2+3(2^2)+4(3^2)+4^2+5(4^2)+6(5^2)+6^2+\dots$
I have noticed that $1^2, 2^2 , 3^2$, etc terms are coming in every $3^{th}$ term . But this observation does not help me to find a solution . Can you help me to find solution of this series ?
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mjb about 10 yearsWhat do you mean by $a.b$?
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Way to infinity about 10 yearsWhere is a.b ? I do not have found a.b terms in my series .
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mjb about 10 yearsWell $(1^2).2$, $(2^2).3$... The question is: what does the period mean?
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Way to infinity about 10 yearsa.b means the the multiplication of two numbers of a and b .
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mjb about 10 yearsI edited your post. Aren't you missing a $3^2$ and a $5^2$ in the sum? What exactly is the series you want to sum? It might be better to write it in the form $a_1=\ldots$, $a_2=\ldots$ etc.
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Way to infinity about 10 yearsNo . I think that $1^2 , 2^2, 4^2 , 6^2,8^2$ terms are in this series .
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gammatester about 10 yearsYour answer seems to ignore that $3^2$ and $5^2$ should not be present in the sum. I think, there cannot be a final answer until an exact specification of the summands is given.
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Natalie about 10 yearstrue - I was hoping that it was just a typo from the asker.