How to do this following "Induction sum"?

n

SUM (-1/2)^j = (2^n+1 + (-1)^n)/ 3. 2^n

j=0

How do we remove the "summation" sign here? Also, how to go about proving it by using "induction"?

Thanks for you reply!

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- May 30th 2008, 08:24 AMrobocop_911Prove by Induction
How to do this following "Induction sum"?

n

SUM (-1/2)^j = (2^n+1 + (-1)^n)/ 3. 2^n

j=0

How do we remove the "summation" sign here? Also, how to go about proving it by using "induction"?

Thanks for you reply! - May 30th 2008, 08:38 AMflyingsquirrel
Hello

To prove this by induction, check that the equality is true for $\displaystyle n=0$. (easy :D)

Then, assuming the equality is true for an integer $\displaystyle n$,

$\displaystyle \begin{aligned}

S_{n+1}=\sum_{j=0}^{n+1}\left(\frac{-1}{2}\right)^j&=\sum_{j=0}^n\left(\frac{-1}{2}\right)^j+\left(\frac{-1}{2}\right)^{n+1}\\

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

\end{aligned}$

Multiply the first fraction by $\displaystyle \frac{2}{2}$ and the second one by $\displaystyle \frac{3}{3}$ so that they can be added.

$\displaystyle S_{n+1}=\frac{2}{2}\cdot\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}+\frac{3}{3}\cdot\left(\frac{-1}{2}\right)^{n+1}=\frac{2^{n+2}+2\cdot(-1)^n}{3\cdot2^{n+1}}+\frac{3\cdot(-1)^{n+1}}{3\cdot 2^{n+1}}=\ldots$

Does it help ? - May 30th 2008, 08:54 AMrobocop_911
Hi, Thanks for you quick reply man!

Anyways, I didn't got how did you "expand"

n

sum (-1/2)^j into (2^n+1+(-1)^n)/3.2^n etc...

j =0

Also, how to get the symbols like you are using? Do we have to write the code for that?

Thanks - May 30th 2008, 09:18 AMflyingsquirrel
I assumed the equality $\displaystyle \sum_{j=0}^n\left(\frac{-1}{2}\right)^j=\frac{2^{n+1}+(-1)^n}{3\cdot 2^n}$ was true for $\displaystyle n$. :D

Quote:

Also, how to get the symbols like you are using? Do we have to write the code for that?

- May 30th 2008, 09:43 AMrobocop_911
- May 30th 2008, 11:16 AMflyingsquirrel
I did this so that the two fractions share the same denominator.

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

\frac{2^{n+1}+(-1)^n}{{\color{blue}3\cdot 2^n}}+\frac{(-1)^{n+1}}{{\color{red}2^{n+1}}}$

As $\displaystyle 3\cdot 2^{n+1}=2\cdot {\color{blue}3\cdot2^n}=3\cdot{\color{red}2^{n+1}}$, we choose $\displaystyle 3\cdot 2^{n+1}$ as the common denominator. To make appear $\displaystyle 3\cdot 2^{n+1}$ at the denominator of the first fraction, we multiply it by $\displaystyle \frac{2}{2}$. For a similar reason, we multiply the second fraction by $\displaystyle \frac{3}{3}$.

Otherwise, one can do this as follows :

$\displaystyle \begin{aligned}

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

\frac{2^{n+1}+(-1)^n}{{3\cdot 2^n}}+\frac{(-1)^{n+1}}{{2^{n+1}}}\\

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

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

&=\ldots

\end{aligned}

$