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

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
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2 Answers

up vote 6 down vote accepted

$A_n=S_n-S_{n-1}=2n+2$. 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 sum {k= 1 to 10} A(2k-1) = sum {k=1 to 10} 4k = 4 * 10 * 11 / 2 = 220 .

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