how to find sum of bC(k-3) +(b+1)C(k-3) + (b+2)C(k-3) + ..... +(b+n-1)C(k-3),where b,n,k, are constants

Question

how to find sum of find sum of bC(k-3) +(b+1)C(k-3) + (b+2)C(k-3) + ..... +(b+n-1)C(k-3) ,where b,n,k, are constants. I am not understanding how to do summations like these..or double summations of such type.Please it will be helpful if somebody answers it. C denotes binomial coefficient, as in nCr =n!/(r!(n-r!))

Answer 1

The first thing you can do is factor out C(k- 3) leaving b+ (b+ 1)+ (b+ 2)+ ...+ (b+ n-1). There are n terms to that so the b's add to nb. All that is left is S(n)= 1+ 2+ 3+ ...+ (n- 2)+ (n- 1). Notice that it is also true that S= (n-1)+ (n-2)+ ...+ 3+ 2+ 1. Also 1+ (n-1)= n, 2+ (n- 2)= n, etc. so if we add those two equations we get 2S on the left and on the right n+ n+ n+ ...+ n, n- 1 times. That is, 2S= n(n- 1) so S= n(n- 1)/2. Putting those together that sum is C(k- 3)(nb)(n(n-1)/2).