### Putnam 1992

**Problem A2**

Let the coefficient of x^{1992} in the power series (1 + x)^{α} = 1 + αx + ... be C(α). Find ∫_{0}^{1} C(-y-1) ∑_{k=1}^{1992}1/(y+k) dy.

**Solution**

Trivial.

The usual Taylor series gives C(-y-1) = (y+1)(y+2) ... (y+1992)/1992! . Hence the integrand is simply the derivative of (y+1)(y+2) ... (y+1992)/1992! , so the integral evaluates to 1993 - 1 = 1992.

Putnam 1992

© John Scholes

jscholes@kalva.demon.co.uk

24 Dec 1998