Question

Let $S_k=\frac{1+2+3+\cdots +k}{k}$ for $k\in N.$ If $S_1^2+S_2^2+\cdots +S_{19}^2=\frac{A}{4},$ then the value of $A$ is

• A. $2869$
• B. $5738$
• C. $1434$
• D. $1436$

Answer 1

The correct option is A $2869$

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

Now, $S_1^2+S_2^2+\cdots +S_{19}^2$
$=\frac{1}{4}(2^2+3^2+4^2+\cdots +{20}^2)$
$=\frac{1}{4}(1^2+2^2+3^2+4^2+\cdots +{20}^2-1^2)$
$=\frac{1}{4}(\frac{20\times 21\times 41}{6}-1)$
$(\because 1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6})$
$=\frac{2869}{4}$
$\therefore A=2869$