sum_{k=1}^5 frac{1^3+2^3+ldots+k^3}{1+3+5+ldo | vedclass

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Question

(A) We know that the sum of the first $k$ cubes is $\sum_{i=1}^k i^3 = \left(\frac{k(k+1)}{2}\right)^2$ and the sum of the first $k$ odd numbers is $\

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(A) We know that the sum of the first $k$ cubes is $\sum_{i=1}^k i^3 = \left(\frac{k(k+1)}{2}\right)^2$ and the sum of the first $k$ odd numbers is $\sum_{i=1}^k (2i-1) = k^2$. Substituting these into the expression: $\sum_{k=1}^5 \frac{\left(\frac{k(k+1)}{2}\right)^2}{k^2} = \sum_{k=1}^5 \frac{k^2(k+1)^2}{4k^2} = \sum_{k=1}^5 \frac{(k+1)^2}{4}$. Expanding the sum for $k=1$ to $5$: $= \frac{1}{4} [2^2 + 3^2 + 4^2 + 5^2 + 6^2] = \frac{1}{4} [4 + 9 + 16 + 25 + 36] = \frac{90}{4} = 22.5$.

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2k-1)}$ is equal to (in $.5$)

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