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Given the sum of first $N$ elements of sequence $A$: $S_{n} = n^{2} + 3n + 4$.
Compute $A_{1} + A_{3} + \cdots + A_{21}$.

I know this problem can be tackled by carefully calculating each value of the sequence. But I wonder what are the better ways to solve it.

Thanks in advance!

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I can write a program for this and it will be solved :), but the program would be much more efficient if I would know the math solution – Omu Aug 4 '10 at 11:47
You should write the first $n$ elements. – Yves Daoust May 10 '15 at 14:28
up vote 8 down vote accepted

\begin{equation*} A_n=S_n-S_{n-1}=2n+2. \end{equation*}

Now one has to find a sum of an arithmetic progression $4+8+\dots+44$.

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Determining a nice form for $A(n)$ is a good start. This is $S(n) - S(n-1) = 2n + 2 $. Now the desired sum is

\begin{equation*} \sum_{k= 1}^{10} A(2k-1) = \sum_{k=1}^{10} 4k = 4 \ast 10 \ast 11 / 2 = 220 .~_{\square} \end{equation*}

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