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!
Hello
To prove this by induction, check that the equality is true for $\displaystyle n=0$. (easy )
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 ?
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 $\displaystyle n$.
Yes, you have to write the code for each formula. See this thread.Also, how to get the symbols like you are using? Do we have to write the code for that?
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}
$