If-C-r-be-the-coefficient-of-x-r-in-1-x-n-then-the-value-of-r-0-n-r-1-2-C-r-is-

Question Number 63831 by gunawan last updated on 10/Jul/19

If C_{r} be the coefficient of x^{r} in {(1 + x)}^{n},

then the value of \underset{r = 0}{\sum^{n}} {(r + 1)}^{2} C_{r} is

Answered by mr W last updated on 10/Jul/19

\underset{r = 0}{\sum^{n}} C_{r} x^{r} = {(1 + x)}^{n}

\underset{r = 0}{\sum^{n}} C_{r} x^{r + 1} = {(1 + x)}^{n} x

\underset{r = 0}{\sum^{n}} (r + 1) C_{r} x^{r} = {(1 + x)}^{n} + n {(1 + x)}^{n - 1} x

\underset{r = 0}{\sum^{n}} (r + 1) C_{r} x^{r + 1} = {(1 + x)}^{n} x + n {(1 + x)}^{n - 1} x^{2}

\underset{r = 0}{\sum^{n}} {(r + 1)}^{2} C_{r} x^{r} = {(1 + x)}^{n} + n {(1 + x)}^{n - 1} x + 2 n {(1 + x)}^{n - 1} x + n (n - 1) {(1 + x)}^{n - 2} x^{2}

l e t x = 1

\underset{r = 0}{\sum^{n}} {(r + 1)}^{2} C_{r} = 2^{n} + n 2^{n - 1} + 2 n 2^{n - 1} + n (n - 1) 2^{n - 2}

= (n + 1) 2^{n} + n (n + 1) 2^{n - 2}

= (n + 1) (n + 4) 2^{n - 2}

Commented by gunawan last updated on 10/Jul/19

wow

thank you Sir

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