Prove that n.1 + (n-1).2 + (n-2).3 ... 3.(n-2) + 2.(n-1) + 1.n = n(n+1)(n+2)/6 By
You can't put n=1 in the L.H.S, when we take p(1) it means the first term i.e. 'n.1' and in the R.H.S n=1 should be put that means p(1) : n.1=1 which is wrong...now can you answer it...please solve it??