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Question Number 54646 by gunawan last updated on 08/Feb/19

$$\mathrm{Such}\mathrm{That}$$$$\mathrm{a}.{}_{n+1}{C}_r=\frac{(n+1).{}_n{C}_r}{(n-r+1)}$$$$\mathrm{b}.{}_n{C}_0{+}_n{C}_2{+}_n{C}_{4...}{=}_n{C}_1{+}_n{C}_3{+}_n{C}_{5...}=2^{n-1}$$$$$$

Answered by Kunal12588 last updated on 08/Feb/19

$$LHS{=}^{n+1}{C}_r=\frac{(n+1)!}{r!(n+1-r)!}$$$$=\frac{(n+1)n!}{r!(n+1-r)(n-r)!}=\frac{n+1}{n-r+1}\times \frac{n!}{r!(n-r)!}$$$$=\frac{(n+1){}^n{C}_r}{(n+1-r)}=RHSproved$$

Commented by gunawan last updated on 08/Feb/19

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

Answered by Kunal12588 last updated on 10/Feb/19

$$b.frombinimialexpansion$$$$0^n={(1-1)}^n$$$$\Rightarrow 0{=}^n{C}_0{-}^n{C}_1{+}^n{C}_2-.....+{(-1)}^n\cdot {}^n{C}_n$$$${\Rightarrow }^n{C}_0{+}^n{C}_2{+}^n{C}_4+....{=}^n{C}_1{+}^n{C}_3{+}^n{C}_5+....eq\left(1\right)$$$$LHSofeq\left(1\right)$$$${}^n{C}_0{+}^n{C}_2{+}^n{C}_4+....$$$${=}^{n-1}{C}_0{+}^{n-1}{C}_1{+}^{n-1}{C}_2{+}^{n-1}{C}_3{+}^{n-1}{C}_4+...$$$$=2^{n-1}proved$$$$Iusedtheidentities\left(1\right)and\left(3\right)from$$$$question54740seecarefully$$

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