Statement -1: sum_{r=0}^{n} (r+1) binom{n}{r} | vedclass

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(D) Consider the sum $S = \sum_{r=0}^{n} (r+1) \binom{n}{r} x^r$.This can be split as $S = \sum_{r=0}^{n} r \binom{n}{r} x^r + \sum_{r=0}^{n} \binom{n

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Statement $-1$ is true,Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

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(D) Consider the sum $S = \sum_{r=0}^{n} (r+1) \binom{n}{r} x^r$.This can be split as $S = \sum_{r=0}^{n} r \binom{n}{r} x^r + \sum_{r=0}^{n} \binom{n}{r} x^r$.Using the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$,we get:$S = \sum_{r=1}^{n} n \binom{n-1}{r-1} x^r + (1+x)^n$$S = nx \sum_{r=1}^{n} \binom{n-1}{r-1} x^{r-1} + (1+x)^n$$S = nx(1+x)^{n-1} + (1+x)^n$.Thus,Statement $-2$ is true.To verify Statement $-1$,substitute $x=1$ into the result of Statement $-2$:$S(1) = n(1)(1+1)^{n-1} + (1+1)^n = n 2^{n-1} + 2^n = 2^{n-1} (n + 2)$.Thus,Statement $-1$ is true and Statement $-2$ is the correct explanation for Statement $-1$.

Statement $-1$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} = (n+2) 2^{n-1}$
Statement $-2$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} x^r = (1+x)^n + nx(1+x)^{n-1}$

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Statement $-1$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} = (n+2) 2^{n-1}$Statement $-2$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} x^r = (1+x)^n + nx(1+x)^{n-1}$