- #1

Lilia

- 48

- 0

## Homework Statement

prove that sum C(2n,k)*k = n*2^(2n-1); k=0,n

## Homework Equations

(x+y)^n = sum [C(n,k)*x^k*y^(n-k)]; k=0,n. // 1

in // 1, when y=1: (1+x)^n = sum [C(n,k)*x^k]; k=0,n // 2

in // 2, when x=1: 2^n = sum [C(n,k)]; k=0,n // 3

derivatives of both sides of (1+x)^n = sum [C(n,k)*x^k]; k=0,n:

n(1+x)^(n-1) = sum [k * C(n,k) * x^(k-1)]; k=0,n // 4

in // 4, when x=1: n*2^(n-1) = sum [k * C(n,k)]; k=0,n // 5

## The Attempt at a Solution

from // 5, i'll have

sum [C(2n,k)] = 2^(2n); k=0,2n

then i tried to divide this by 2:

sum [C(2n,k)] = 2^(2n)/2 = 2^(2n-1); k=0,n

but then i realized this can't be correct because there are

n+1 numbers in the range 0-n,

2n+1 numbers in the range 0-2n

so i can't really divide 2n+1 by two and get n+1