5

I'm trying to solve this problem about mathematical induction but every time I try to solve it I end up with an incorrect answer. The problem is as follows:

Prove by induction : $$\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $$

my work: first I prove that it holds for n=1: $\frac 12 = 1- \frac {1}{2^1} = \frac 12$

Now assume it holds for n=k, prove for n=k+1 the sum is equal to $1 - \frac{1}{2^{k+1}}$

then $\frac 12 + \frac 14 +...+ \frac {1}{2^k} + \frac {1}{2^{k+1}} = 1-\frac {1}{2^k} + \frac {1}{2^{k+1}} = 1 - \frac {2^{k+1}+2^k}{2^{k+1}2^k} = 1- \frac {2(2^k)+2^k}{2(2^k)2^k} = 1 - \frac {3(2^k)}{2(2^k)2^k} = 1- \frac {3}{2(2^k)} = 1- \frac {3}{2^{k+1}} $

I would very much appreciate if at least someone can tell me where I'm wrong so I could try to solve it by myself. I think the mistake is at factoring $2^{k+1}+2^k$

Mr. Nicolas
  • 169

2 Answers2

5

Be careful with the minus sign. $$-\frac{1}{2^k} + \frac{1}{2^{k+1}} = \frac{-2^{k+1} + 2^k}{2^k 2^{k+1}} = \frac{-2 + 1}{2^{k+1}} = -\frac{1}{2^{k+1}}.$$

angryavian
  • 89,882
4

You went wrong when you combined the fractions. You should have $$1- \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{2^{k+1} - 2^k}{2^{k+1}2^k} = 1 - \frac{2 - 1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}$$

PTrivedi
  • 1,001