Statement - 1 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
Statement - 2 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $$
A
Statement - 1 is false, Statement - 2 is true
B
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
C
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
D
Statement - 1 is true, Statement - 2 is false
Explanation
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,
$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
So Option B is correct.