If f (x) is a polynomial of degree n such that f (0) = 0, f(1)=12,…,f(n)=n(n+1), then the value of f (n + 1) is

(x+1)f(x)−x is a polynomial of degree n + 1. Therefore,(x+1)f(x)−x=k(x)[x−1][x−2]…[x−n]         (1)or [n+2]f(n+1)−(n+1)=k[(n+1)!]Also 1=k(−1)(−2)…((−n−1)) [ Putting x=−1 in (1) ]or 1=k(−1)n+1(n+1)!or (n+2)f(n+1)−(n+1)=(−1)n+1Hence, f(n+1)=1, if n is odd, and nn+2, if n is even.