Sequences & Series Question - None

Question

Given $S_k = \dfrac{1+2+3+\cdots+k}{k}$ and $\displaystyle\sum_{k=1}^{10} S_k^2 = \dfrac{5}{12}A$. Find the value of $A$.

Accepted Answer

Step 1: Simplify $S_k$:

$S_k = \frac{1+2+3+\cdots+k}{k} = \frac{\frac{k(k+1)}{2}}{k} = \frac{k+1}{2}$

Step 2: Find $S_k^2$:

$S_k^2 = \left(\frac{k+1}{2}\right)^2 = \frac{(k+1)^2}{4}$

Step 3: Calculate $\sum_{k=1}^{10} S_k^2$:

$\sum_{k=1}^{10} S_k^2 = \sum_{k=1}^{10} \frac{(k+1)^2}{4} = \frac{1}{4}\sum_{k=1}^{10} (k+1)^2$

$= \frac{1}{4}\sum_{j=2}^{11} j^2$ (substituting $j = k+1$)

$= \frac{1}{4}\left(\sum_{j=1}^{11} j^2 - 1\right)$

Step 4: Use the formula $\sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}$:

$\sum_{j=1}^{11} j^2 = \frac{11 \cdot 12 \cdot 23}{6} = \frac{3036}{6} = 506$

Step 5: Substitute back:

$\sum_{k=1}^{10} S_k^2 = \frac{1}{4}(506 - 1) = \frac{505}{4}$

Step 6: Solve for $A$:

$\frac{505}{4} = \frac{5}{12}A$

$A = \frac{505}{4} \cdot \frac{12}{5} = \frac{505 \cdot 12}{20} = \frac{6060}{20} = 303$

∴ Answer: 303

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