Hi I hope someone can help me with this problem:
Prove the following formula by induction:
1+(1/2)+(1/4)+(1/8)+...+1/(2^n) = 2 - 1/(2^n)
Thanks in advance
A) Base case n=0, then the left hand side is 1, and the right hnd side is 1.
B) Suppose this true for some $\displaystyle n=k \in \mathbb{N_+}$
Then:
$\displaystyle
[1+1/2+1/4+1/8+...+1/2^{k}]+1/2^{k+1} = [2 - 1/2^k]+1/2^{k+1} = 2-1/2^{k+1}
$
hence it is true for k+1.
Thus combining A), and B) we have proven that it is true for all $\displaystyle n \in \mathbb{N}$ by mathematical induction
RonL
n = 0
$\displaystyle 1 = 2 - \frac{1}{2^0} = 2 - 1 = 1$ (Check!)
Now assume this is true for some n = k. We wish to show that it is true for n = k + 1.
So assume:
$\displaystyle 1 + \frac{1}{2} + ~ ... ~ + \frac{1}{2^k} = 2 - \frac{1}{2^k}$
is true.
$\displaystyle 1 + \frac{1}{2} + ~ ... ~ + \frac{1}{2^k} + \frac{1}{2^{k + 1}}$
$\displaystyle = \left ( 2 - \frac{1}{2^k} \right ) + \frac{1}{2^{k + 1}}$
$\displaystyle = 2 - \left ( \frac{1}{2^k} - \frac{1}{2^{k + 1}} \right ) $
$\displaystyle = 2 - \left ( \frac{1}{2^k} - \frac{1}{2} \cdot \frac{1}{2^k} \right ) $
$\displaystyle = 2 - \left ( 1 - \frac{1}{2} \right ) \frac{1}{2^k} $
$\displaystyle = 2 - \frac{1}{2} \cdot \frac{1}{2^k} $
$\displaystyle = 2 - \frac{1}{2^{k + 1}}$
as we needed to show.
-Dan