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If p+q=1, then show that sum(r=0)^n r^2^...

If `p+q=1,` then show that `sum_(r=0)^n r^2^n C_rp^r q^(n-r)=n p q+n^2p^2dot`

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statement 1: Let p_1,p_2,...,p_n and x be distinct real number such that (sum_(r=1)^(n-1)p_r^2)x^2+2(sum_(r=1)^(n-1)p_r p_(r+1))x+sum_(r=2)^n p_r^2 lt=0 then p_1,p_2,...,p_n are in G.P. and when a_1^2+a_2^2+a_3^2+...+a_n^2=0,a_1=a_2=a_3=...=a_n=0 Statement 2 : If p_2/p_1=p_3/p_2=....=p_n/p_(n-1), then p_1,p_2,...,p_n are in G.P.

Consider a G.P. with first term (1+x)^(n) , |x| lt 1 , common ratio (1+x)/(2) and number of terms (n+1) . Let 'S' be sum of all the terms of the G.P. , then sum_(r=0)^(n)"^(n+r)C_(r )((1)/(2))^(r ) equals

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Prove that sum_(r=1)^n(-1)^(r-1)(1+1/2+1/3++1/r)^n C_r=1/n .