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

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(C) We have $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $.Using the identity $r^2 = r(r-1) + r$,we get:$= \sum\limits_{r = 0}^n {[r(r -

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$nx(nx + y)$

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(C) We have $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $.Using the identity $r^2 = r(r-1) + r$,we get:$= \sum\limits_{r = 0}^n {[r(r - 1) + r]{\,^n}} {C_r}{x^r}{y^{n - r}}$$= \sum\limits_{r = 2}^n {r(r - 1){\,^n}} {C_r}{x^r}{y^{n - r}} + \sum\limits_{r = 1}^n {{r^n}{C_r}{x^r}{y^{n - r}}}$$= \sum\limits_{r = 2}^n {n(n - 1){\,^{n - 2}}{C_{r - 2}}{x^r}{y^{n - r}}} + \sum\limits_{r = 1}^n {n{\,^{n - 1}}{C_{r - 1}}{x^r}{y^{n - r}}}$$= n(n - 1){x^2}\sum\limits_{r = 2}^n {{\,^{n - 2}}{C_{r - 2}}{x^{r - 2}}{y^{(n - 2) - (r - 2)}}} + nx\sum\limits_{r = 1}^n {{\,^{n - 1}}{C_{r - 1}}{x^{r - 1}}{y^{(n - 1) - (r - 1)}}}$$= n(n - 1){x^2}{(x + y)^{n - 2}} + nx{(x + y)^{n - 1}}$Since $x + y = 1$,this simplifies to:$= n(n - 1){x^2} + nx$$= n^2x^2 - nx^2 + nx = nx(nx - x + 1)$Since $1 - x = y$,we have:$= nx(nx + y)$

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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