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Question Number 117649 by Ar Brandon last updated on 13/Oct/20

$$\mathrm{Let}\mathrm{P}\left(x\right)\mathrm{be}\mathrm{a}\mathrm{polynomial}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{n}\mathrm{such}\mathrm{that}$$$$\mathrm{P}\left(k\right)=\frac{k}{k+1}\mathrm{for}k=0,1,2,...,\mathrm{n}.\mathrm{Then}\mathrm{P}(\mathrm{n}+1)\mathrm{is}\mathrm{equal}\mathrm{to}$$$$\left(\mathrm{A}\right)-1\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\left(\mathrm{B}\right)1\mathrm{if}n\mathrm{is}\mathrm{odd}$$$$\left(\mathrm{C}\right)\frac{n}{n+2}\mathrm{if}n\mathrm{is}\mathrm{even}\left(\mathrm{D}\right)\frac{n}{n+2}\mathrm{if}n\mathrm{is}\mathrm{odd}$$$$\mathrm{Which}\mathrm{among}\mathrm{the}\mathrm{four}\mathrm{proposals}\mathrm{is}/\mathrm{are}\mathrm{correct}?$$

Answered by prakash jain last updated on 13/Oct/20

$$Q\left(k\right)=(k+1)P\left(k\right)-k.....\left(A\right)$$$$Q\left(0\right)=0$$$$Q\left(1\right),....Q\left(n\right)=0$$$$Q\left(k\right)\mathrm{has}\mathrm{zeros}\mathrm{at}0,1,2,...,n$$$$Q\left(k\right)=ck(k-1)(k-2)....(k-n)$$$$From\left(A\right)Q(-1)=1$$$$1=c(-1)(-2)....(-1-n)$$$$c=\frac{{(-1)}^{n+1}}{(n+1)!}$$$$Q(n+1)=\frac{{(-1)}^{n+1}}{(n+1)!}(n+1)n(n-1)...1$$$$={(-1)}^{n+1}$$$$from\left(A\right)$$$$P(n+1)=\frac{Q(n+1)+(n+1)}{n+2}$$$$=\frac{(n+1)+{(-1)}^{n+1}}{n+2}$$

Commented by Ar Brandon last updated on 13/Oct/20

Thank You Sir

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