mathematical induction (2)

Let P(n) : [2 / (x - 2 ) ] - [ 2/x + (2^ 2)/(x^ 2) + . . . + (2^n) / (x^n) ] = [2 ^ ( n + 1 )] / [ (x^ n) ( x - 2 ) ] for all natural number n .
For n = 1,
L.H.S = [2/(x-2)] - [2/x] = (2x-2x+4)/x(x-2) = 4/ x(x-2)
R.H.S = 2^2/x(x-2) = 4/x(x-2) = L.H.S
P(1) holds

Assume P(k) holds for some natural number k, then
P(k) : [2 / (x - 2 ) ] - [ 2/x + (2^ 2)/(x^ 2) + . . . + (2^k) / (x^k) ] = [2 ^ ( k + 1 )] / [ (x^ k) ( x - 2 ) ]

for n=k+1,
L.H.S = [2 / (x - 2 ) ] - [ 2/x + (2^ 2)/(x^ 2) + . . . + [2^(k+1)] / [x^(k+1)]
= [2 ^ ( k + 1 )] / [ (x^ k) ( x - 2 ) ] - [2^(k+1)] / [x^(k+1)]
= { x[2^(k+1)]- [2^(k+1)](x-2) } / [x^(k+1)(x-2)]

= 2^(k+2)/x^(k+1)(x-2)

R.H.S = 2^[(k+1)+1] / [ x^(k+1)(x-2) = 2^(k+2)/ x^(k+1)(x-2) = L.H.S


P(k+1) holds



Hence, by Mathematic induction, P(n) holds for all natural number n