Question Number 56139 by gunawan last updated on 11/Mar/19
$$\mathrm{If}\:{x}+{y}=\mathrm{1},\:\mathrm{then}\:\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:{r}^{\mathrm{2}} \:\:^{{n}} {C}_{{r}} \:{x}^{{r}} \:{y}^{{n}−{r}} \:\mathrm{equals} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
$$\left({y}+{x}\right)^{{n}} =\mathrm{1} \\ $$$${nc}_{\mathrm{0}} {y}^{{n}−\mathrm{0}} {x}^{\mathrm{0}} +{nc}_{\mathrm{1}} {y}^{{n}−\mathrm{1}} {x}^{\mathrm{1}} +{nc}_{\mathrm{2}} {y}^{{n}−\mathrm{2}} {x}^{\mathrm{2}} +...+{nc}_{{n}} {y}^{{n}−{n}} {x}^{{n}} \\ $$$$\left({y}+{x}\right)^{{n}} =\mathrm{1}=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nc}_{{r}} {y}^{{n}−{r}} {x}^{{r}} \\ $$$${wait}...{pls}... \\ $$$$ \\ $$