Let S_k = 1+2+3++kk.

Question

Let $S_k = \dfrac{1+2+3+\cdots+k}{k}$. If $S_1^2+S_2^2+\cdots+S_{10}^2 = \dfrac{5}{12}A$, then $A$ is equal to:

Accepted Answer

Sk = $${{K + 1} \over 2}$$

$$\sum {S_k^2}  = {5 \over {12}}$$ A

$$\sum\limits_{K = 1}^{10} {{{\left( {{{K + 1} \over 2}} \right)}^2}}  = {{{2^2} + {3^2} +  -  -  + {{11}^2}} \over 4} = {5 \over {12}}$$ A

$${{11 \times 12 \times 23} \over 6} - 1 = {5 \over 3}$$ A

505  $$ = {5 \over 3}$$ A,   A = 303

Solutions · 1