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Let $S_{k}=\frac{1+2+3+\cdots +k}{k}.$ If ${S_{1}}^{2}+{S_{2}}^{2}+\cdots +{S_{10}}^{2}=\frac{5}{12}A,$ then A is equal to :
Option 1)
$283$
Option 2)
$301$
Option 3)
$156$
Option 4)
$303$

Answers (1)

best_answer

Summation of series of natural numbers -

$\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )$

- wherein

Sum of first n natural numbers

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

$S_{k}=\frac{k(k+1)}{2k}=\frac{k+1}{2}$

$=>\sum_{i=1}^{10}(\frac{k+1}{2})^{2}=\frac{5}{12}A$

$=>2^{2}+3^{2}+4^{2}+....................+11^{2}=\frac{5}{12}A$

Since $\frac{n(n+1)(2n+1)}{6}=\sum_{i=1}^{n}m^{2}$

$=>\frac{11\times 12\times 23}{6}-1=\frac{5}{12}A$

$=>A=303$

Option 1)

$283$

Option 2)

$301$

Option 3)

$156$

Option 4)
$303$

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