1^2+3+5^2+7+9^2+11+ up to 40 terms is equal to:

Question

$1^2+3+5^2+7+9^2+11+\cdots$ up to 40 terms is equal to:

Accepted Answer

Grouping the series into the squared terms $(1^2,5^2,9^2,\ldots$, 20 terms$)$ and the linear terms $(3,7,11,\ldots$, 20 terms$)$:

$$\sum_{k=1}^{20}(4k-3)^2 = 16\sum k^2 - 24\sum k + 9(20), \qquad \sum_{k=1}^{20}\big(6+(k-1)4\big) = 10\times82 = 820.$$

$$16(2870) - 24(210) + 180 + 820 = 45920-5040+1000 = 41880.$$

Solutions · 1