Prove that n x 2n 1xn x 0Solutiona sigma x 0n nCx nC0

Prove that (n x) = 2n. (-1)x(n x) = 0.

Solution

a)
sigma (x = 0->n ) nCx
= nC0 + nC1 + nC2 + nC3 + .....
= 1+ n + n(n-1)/2! +n(n-1)(n-2)/3! + ..... ------------>1

Also from binomial expansion ,
(1+x)^n = 1 +nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + ....... ----------->2

Putting the value x=1 in Right Hand Side of equation 2 gives :

(1+1)^n = 1 +n + n(n-1)/2! + n(n-1)(n-2)/3! + .......
=> 2^n = 1 +n + n(n-1)/2! + n(n-1)(n-2)/3! + .......
=>2^n = sigma (x = 0->n ) nCx [From equation 1]
or,
sigma (x = 0->n ) nCx = 2^n

Proved

b)
sigma (x = 0->n ) (-1)^x nCx
= (-1)^0 * nC0 + (-1)^1 * nC1 + (-1)^2 * nC2 + (-1)^3 * nC3 + .....
= nC0 - nC1 + nC2 - nC3 + .....
= 1- n + n(n-1)/2! - n(n-1)(n-2)/3! + ..... ------------>1

Also from binomial expansion ,
(1-x)^n = 1 - nx + n(n-1)x^2/2! - n(n-1)(n-2)x^3/3! + ....... ----------->2

Putting the value x=1 in Right Hand Side of equation 2 gives :

(1-1)^n = 1 - n + n(n-1)/2! - n(n-1)(n-2)/3! + .......
=> 0 = 1 - n + n(n-1)/2! - n(n-1)(n-2)/3! + .......
=>0 = sigma (x = 0->n ) (-1)^x nCx [From equation 1]
or,
sigma (x = 0->n ) (-1)^x nCx = 0

Proved