# using mathematical induction to prove..

• Mar 11th 2009, 08:19 PM
jvignacio
using mathematical induction to prove..
hi guys having trouble proving this result.

$
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},$

for all $n \in \mathbb{N}$

Any help much appreciated.. thanks
• Mar 11th 2009, 09:14 PM
TheEmptySet
Quote:

Originally Posted by jvignacio
hi guys having trouble proving this result.

$
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},$

for all $n \in \mathbb{N}$

Any help much appreciated.. thanks

First show the base case is true n=1

$
\sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}=\frac{1}{2}?=\frac{2}{3} + \frac{1}{3}\frac{(-1)^1}{2^1}=\frac{1}{2},$

so it checks. Now assume it is true for n

$
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},$

no show N+1

$
\sum_{k=0}^{n+1}(-\frac{1}{2})^k=\left(-\frac{1}{2}\right)^{n+1}+\sum_{k=0}^{n}(-\frac{1}{2})^k$

Now by the induction hypothesis we get

$\left(-\frac{1}{2}\right)^{n+1} +\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n}$

$\frac{2}{3}+\left(-\frac{1}{2}\right)^{n+1}+\frac{1}{3}\frac{(-1)^n}{2^n}=\frac{2}{3}+\left(-\frac{1}{2}\right)\frac{(-1)^{n}}{2^n}+\frac{1}{3}\frac{(-1)^n}{2^n}$

$\frac{2}{3}+\left[\left( -\frac{1}{2}\right) +\frac{1}{3}\right]\frac{(-1)^n}{2^n}=\frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}$

$\frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}\left( \frac{-1}{2}\right)\left( \frac{2}{-1}\right)=\frac{2}{3}+\left( \frac{2}{3}\right) \frac{(-1)^{n+1}}{2^{n+1}}$

This shows n+1 so we are done
• Mar 11th 2009, 09:23 PM
jvignacio
Quote:

Originally Posted by TheEmptySet
First show the base case is true n=1

$
\sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}=\frac{1}{2}?=\frac{2}{3} + \frac{1}{3}\frac{(-1)^1}{2^1}=\frac{1}{2},$

so it checks. Now assume it is true for n

$
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},$

no show N+1

$
\sum_{k=0}^{n+1}(-\frac{1}{2})^k=\left(-\frac{1}{2}\right)^{n+1}+\sum_{k=0}^{n}(-\frac{1}{2})^k$

Now by the induction hypothesis we get

$\left(-\frac{1}{2}\right)^{n+1} +\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n}$

$\frac{2}{3}+\left(-\frac{1}{2}\right)^{n+1}+\frac{1}{3}\frac{(-1)^n}{2^n}=\frac{2}{3}+\left(-\frac{1}{2}\right)\frac{(-1)^{n}}{2^n}+\frac{1}{3}\frac{(-1)^n}{2^n}$

$\frac{2}{3}+\left[\left( -\frac{1}{2}\right) +\frac{1}{3}\right]\frac{(-1)^n}{2^n}=\frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}$

$\frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}\left( \frac{-1}{2}\right)\left( \frac{2}{-1}\right)=\frac{2}{3}+\left( \frac{2}{3}\right) \frac{(-1)^{n+1}}{2^{n+1}}$

This shows n+1 so we are done

very well explained. much appreciated
• Mar 11th 2009, 09:26 PM
jvignacio
Quote:

Originally Posted by TheEmptySet
First show the base case is true n=1

$\sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}$

quick question how did u get $1-\frac{1}{2}$ the 1 - ?
• Mar 12th 2009, 09:28 AM
GaloisTheory1
Quote:

Originally Posted by jvignacio
$\sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}$

quick question how did u get $1-\frac{1}{2}$ the 1 - ?

plug in $k=0$ and $k=1$ into the sum, see what you get.