using the induction method Prove that 121 222 323 n2n 1

using the induction method

Prove that 1/2^1 + 2/2^2 + 3/2^3 + ... n/2^n 1. Using the induction method

Base Case (n=1)

LHS = 1/(2)^1 = 1/2

RHS = 2

Hence LHS < RHS, therefore base case holds true

Inductive Step (n=k)

We assume that the given thing is valid for n=k

(1/2^1 + 2/2^2 + ... + k/2^(k)) < 2

Las Step (n=k+1)

We need to prove that LHS is less than 2

LHS = (1/2^1 + 2/2^2 + .... + k/2^k + (k+1)/2^(k+1))

=> (1/2^1 + 2/2^2 + ... + k/2^k + k/2^(k+1) + 1/2^(k+1)

=> (1/2^1 + 2/2^2 + ... + (k-1)/2^(k-1) + (3k+1)/2^(k+1) ) < 2

Since the all the first k terms are less than 2 for this equation

Hence we prove the given inequality using mathematical induction