Let S_1=_j=1^10j(j-1),^10C_j, S_2=_j=1^10j,^10C_j, and S_3=_j=1^10j^2,^10C_j.

Question

Let $S_1=\displaystyle\sum_{j=1}^{10}j(j-1)\,{}^{10}C_j$, $S_2=\displaystyle\sum_{j=1}^{10}j\,{}^{10}C_j$, and $S_3=\displaystyle\sum_{j=1}^{10}j^2\,{}^{10}C_j$.

Statement -1: $S_3=55	imes2^9$

Statement -2: $S_1=90	imes2^8$ and $S_2=10	imes2^8$

Accepted Answer

Note :

$$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ = $= n{.2^{n - 1}}$

$$\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}}  = n\left( {n + 1} \right){2^{n - 2}}$$

Given that,

$${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$

=$$\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}$$

= 10$$ \times $$11$$ \times $$ ${2^{10 - 2}}$- 10$ \times $$ ${2\^{10 - 1}}$$

= 10$$ \times $$ ${2^{8}}$(11 - 2)

= 10$ \times $9$ \times $$ ${2\^{8}}$$

= 90$$ \times $$ ${2^{8}}$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$= 10$ \times $$ ${2\^{10-1}}$$

= 10$$ \times $$ ${2^{9}}$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$= 10$ \times $11$ \times $$ ${2\^{10-2}}$$

= $${{110} \over 2} \times {2^9}$$

= 55 $$ \times $$ ${2^9}$

Solutions · 1