Statement - 1 : [_r = 0^n ( r + 1 ),^nC_r = ( n + 2 )2^n - 1. ] Statement - 2 : [_r = 0^n …

Question

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

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

Accepted Answer

Check Statement - 1

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

= $$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ + $$\sum\limits_{r = 0}^n {{}^n{C_r}} $$

= $$\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}} $$ $+ {2^n}$

= $$n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} $$ $+ {2^n}$

= $$n \times {2^{n - 1}}$$ $ + {2^n}$=${2^{n - 1}}\left[ {n + 2} \right]$Check Statement  2 :$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}$=$\sum\limits_{r = 0}^n r .{}^n{C_r}.{x^r}$+$\sum\limits_{r = 0}^n {{}^n{C_r}.{x^r}} $=$n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^r}} $$ + {\left( {1 + x} \right)^n}$=$nx\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^{r - 1}}}  + {\left( {1 + x} \right)^n}$=$nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n}$$

Substitude x = 1 in the statement 2 and we get,

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

Solutions · 1