Question
Statement - 1 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{\,^n}{C_r}} = \left( {n + 2} \right){2^{n - 1}}.$$
Statement - 2 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{\,^n}{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $$
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
Answer :
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
Solution :
We have
$$\eqalign{
& \sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{\,^n}{C_r}{x^r} = } \sum\limits_{r = 0}^n {r.{\,^n}{C_r}{x^r} + \sum\limits_{r = 0}^n {^n{C_r}{x^r}} } \cr
& = \sum\limits_{r = 1}^n {r.\,\frac{n}{r}{\,^{n - 1}}{C_{r - 1}}{x^r} + {{\left( {1 + x} \right)}^n}} \cr
& = nx\sum\limits_{r = 1}^n {^{n - 1}{C_{r - 1}}{x^{r - 1}} + {{\left( {1 + x} \right)}^n}} \cr
& = nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n} = {\text{RHS}} \cr} $$
∴ Statement - 2 is correct.
Putting $$x = 1,$$ we get
$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{\,^n}{C_r} = n.\,{2^{n - 1}}} + {2^n} = \left( {n + 2} \right){.2^{n - 1}}.$$
∴ Statement - 1 is also true and statement - 2 is a correct explanation for statement - 1.