If x + y = 1, then sumlimits_{r = 0}^n {{r^2} | vedclass

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Question

(c) We have

$\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $

$ = \sum\limits_{r = 0}^n {[r(r - 1) + r]{\,^n}} {C_r}{x^r}{y^{n - r}}$&

Vedclass · Accepted Answer

$nx(nx + y)$

Vedclass · Answer

(c) We have

$\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $

$ = \sum\limits_{r = 0}^n {[r(r - 1) + r]{\,^n}} {C_r}{x^r}{y^{n - r}}$ 

$ = \sum\limits_{r = 0}^n {r(r - 1){\,^n}} {C_r}{x^r}{y^{n - r}} + \sum\limits_{r = 0}^n {{r^n}{C_r}{x^r}{y^{n - r}}} $ 

$ = \sum\limits_{r = 2}^{n - 2} {r(r - 1)\frac{n}{r}.\frac{{n - 1}}{{r - 1}}{\,^{n - 2}}{C_{r - 2}}{x^2}{x^{r - 2}}{y^{n - r}}} $ $ + \sum\limits_{r = 1}^{n - 1} {r\frac{n}{r}{\,^{n - 1}}{C_{r - 1}}x\,\,{x^{r - 1}}{y^{n - r}}} $

$ = n(n - 1){x^2}\sum\limits_{r = 2}^{n - 2} {{\,^{n - 2}}{C_{r - 2}}{x^{r - 2}}{y^{(n - 2) - (r - 2)}}} $ $+ nx\sum\limits_{r = 2}^{n - 1} {{\,^{n - 1}}{C_{r - 1}}{X^{r - 1}}{y^{(n - 1) - (r - 1)}}}$

$ = n(n - 1){x^2}{(x + y)^{n - 2}} + nx{(x + y)^{n - 1}}$

$ = n(n - 1){x^2} + nx,\,\,\,\,\,(\because x + y = 1)$ 

$ = nx(nx - x + 1) = nx(nx + y)\,,\,\,(\because x + y = 1)$

If $x + y = 1$, then $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $ equals

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