Let Q(x)=(x+1)P(x)-x.. Then Q has roots 0, 1, ..., n and Q is of degree n+1. Therefore, Q(x)=Cx(x-1)...(x-n).. Also note Q(-1)=1, get C=(-1)^(n+1) 1/(n+1)!. Q(n+1)=(-1)^(n+1). P(n+1)=(n+1+(-1)^(n+1))/(n+2). |
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