Aspire Faculty ID #18680 · Topic: NIMCET 2026 · Just now
NIMCET 2026

Let $A_k$ be the arithmetic mean of squares of $k$ natural numbers. If $\sum_{k=1}^{n}(6A_k-3k)=31$, find the value of $n$.

Solution

$A_k=\frac{1^2+2^2+3^2+\cdots+k^2}{k}$ 
$A_k=\frac{\frac{k(k+1)(2k+1)}{6}}{k}$ 
$A_k=\frac{(k+1)(2k+1)}{6}$ 
Now, $6A_k-3k=(k+1)(2k+1)-3k$ 
$=2k^2+3k+1-3k$ $=2k^2+1$ 
Given, 
$\sum_{k=1}^{n}(6A_k-3k)=31$ 
$\sum_{k=1}^{n}(2k^2+1)=31$ 
For $n=3$, 
$(2\cdot1^2+1)+(2\cdot2^2+1)+(2\cdot3^2+1)$ 
$=3+9+19=31$ Therefore, 
$n=3$
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