Question

State and prove bionomial theorm

Answer 1

For any positive integer n,

(x+y)n=nk=0n k  xn−kyk  

Proof by Induction:

For n=1,

(x+y)1=x+y=1 0  x1−0y0+1 1  x1−1y1=1k=01 k  x1−kyk  

Suppose (x+y)n−1=n−1k=0n−1 k  x(n−1)−kyk  

Consider (x+y)n.

(x+y)n = = = = = =   = = (x+y)(x+y)n−1 (x+y)n−1k=0n−1 k  x(n−1)−kyk n−1k=0n−1 k  xn−kyk+n−1j=0n−1 j  x(n−1)−jyj+1 n−1k=0n−1 k  xn−kyk+n−1j=0n−1 (j+1)−1  xn−(j+1)yj+1 n−1k=0n−1 k  xn−kyk+nk=1n−1 k−1  xn−kyk nk=0n−1 k  xn−kyk−n−1 n  x0yn +nk=0n−1 k−1  xn−kyk−n−1 −1  xny0 nk=0n−1 k  +n−1 k−1  xn−kyk nk=0n k  xn−kyk  

Answer 2

For any positive integer n,

(x+y)n=nk=0n k  xn−kyk  

Proof by Induction:

For n=1,

(x+y)1=x+y=1 0  x1−0y0+1 1  x1−1y1=1k=01 k  x1−kyk  

Suppose (x+y)n−1=n−1k=0n−1 k  x(n−1)−kyk  

Consider (x+y)n.

(x+y)n = = = = = =   = = (x+y)(x+y)n−1 (x+y)n−1k=0n−1 k  x(n−1)−kyk n−1k=0n−1 k  xn−kyk+n−1j=0n−1 j  x(n−1)−jyj+1 n−1k=0n−1 k  xn−kyk+n−1j=0n−1 (j+1)−1  xn−(j+1)yj+1 n−1k=0n−1 k  xn−kyk+nk=1n−1 k−1  xn−kyk nk=0n−1 k  xn−kyk−n−1 n  x0yn +nk=0n−1 k−1  xn−kyk−n−1 −1  xny0 nk=0n−1 k  +n−1 k−1  xn−kyk nk=0n k  xn−kyk