| 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 true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is false.
C) Statement-1 is false, Statement-2 is true.
D) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Correct Answer: D
Solution :
\[\sum\limits_{r=0}^{n}{{{\left( r+1 \right)}^{n}}{{C}_{r}}{{x}^{r}}=}\sum\limits_{r=0}^{n}{r.{{\,}^{n}}{{C}_{r}}{{x}^{r}}+\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}.\,{{x}^{n}}}}\] \[=nx\sum\limits_{r=0}^{n}{^{n-1}{{C}_{r-1}}{{x}^{r-1}}+}\] \[\sum\limits_{r=0}^{n}{^{n}\,{{C}_{r}}{{x}^{r}}=nx{{\left( 1+x \right)}^{n-1}}+{{\left( 1+x \right)}^{n}}}\] ?. (i) Statement-2 is true. Putting \[x=1\] in (i), we get \[\sum\limits_{r=0}^{n}{\left( r+1 \right).{{\,}^{n}}{{C}_{r}}=\left( n+2 \right).\,{{2}^{n-1}}}\]. Statement-1 is also true.
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