Expand (x + 2x2 + 3x3 + ... + nxn)2 and add the coefficients of xn+1 through x2n. Show that the result is n(n+1)(5n2 + 5n + 2)/24.
Solution
Let the coefficient of xm be am. Multiplying out, we find:
an+1 = 1.n + 2(n-1) + 3(n-2) + ... + (n-2)3 + (n-1)2 + n.1 an+2 = 2.n + 3(n-1) + 4(n-2) + ... + (n-1)3 + n.2 an+3 = 3.n + 4(n-1) + 5(n-2) + ... + n.3 ... a2n-1 =(n-1)n +n(n-1) a2n = n.nAdding the columns, we get (1+ ... + n)n + (2+...+n)(n-1) + (3+...+n)(n-2) + ... + (n-1+n)2 + n.1. Doubling, we get (n+1)n2 + (n+2)(n-1)2 + (n+3)(n-2)2 + ... + (2n-1)22 + 2n·12 = n(12 + ... + n2) + (1·n2 + 2(n-1)2 + 3(n-2)2 + ... + n.12).
It is well-known that 12 + ... + n2 = n(n+1)(2n+1)/6. We claim that 1.n2 + 2(n-1)2 + ... + n.12 = n(n+1)2(n+2)/12. The result then follows immediately, since ( n2(n+1)(2n+1)/6 + n(n+1)2(n+2)/12 )/2 = n(n+1) ( 2n(2n+1) + (n+1)(n+2) )/24 = n(n+1)(5n2+5n+2)/24.
It is also well-known that 13 + 23 + ... + n3 = n2(n+1)2/4. [If you forget the sums of the powers it is easiest to start with ∑ r(r+1)(r+2) ... (r+m) = n(n+1)...(n+m+1)/(m+2) which is easy to remember and trivial to prove (by induction).]. Adding n·n2 + (n-1)·(n-1)2 + ... + 1·12 to 1·n2 + 2(n-1)2 + ... + n·12 we get (n+1)(n2 + (n-1)2 + ... + 12) = n(n+1)2(2n+1)/6. Hence 1.n2 + 2(n-1)2 + ... + n.12 = n(n+1)2(2n+1)/6 - n2(n+1)2/4 = 1/12 n(n+1)2(4n+2 - 3n) = n(n+1)2(n+2)/12, as claimed.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002