Ok, so I'm not very good with these proving by induction thingies. Need a little help please.
How do I prove
$$\sum_{j=0}^n\left(-\frac12\right)^j=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}$$
when $n$ is a non-negative integer.
I got the basis step and such down, but I'm pretty bad with exponents so I am having a difficult time in the induction step.
Your induction hypothesis is that
$$\sum_{j=0}^n\left(-\frac12\right)^j=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}\;,\tag{1}$$
and you want to prove that
$$\sum_{j=0}^{n+1}\left(-\frac12\right)^j=\frac{2^{n+1}+(-1)^{n+1}}{3\cdot2^{n+1}}\;.\tag{2}$$
The natural thing to do is to split the sum in $(2)$ into the sum in $(1)$ and the new, extra term and apply the induction hypothesis to get
$$\begin{align*} \sum_{j=0}^{n+1}\left(-\frac12\right)^j&=\sum_{j=0}^n\left(-\frac12\right)^j+\left(-\frac12\right)^{n+1}\\\\ &=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}+\left(-\frac12\right)^{n+1}\\\\ &=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}+\frac{(-1)^{n+1}}{2^{n+1}}\;. \end{align*}$$
Now you clearly want to put the two fractions over a common denominator to combine them:
$$\begin{align*} \frac{2^{n+1}+(-1)^n}{3\cdot2^n}+\frac{(-1)^{n+1}}{2^{n+1}}&=\frac{2\left(2^{n+1}+(-1)^n\right)+3(-1)^{n+1}}{3\cdot2^{n+1}}\\\\ &=\frac{2^{n+2}+2(-1)^n+3(-1)^{n+1}}{3\cdot2^{n+1}}\;. \end{align*}$$
Now there’s just one more small step, which I’ll leave to you; remember that $(-1)^n=-(-1)^{n+1}$. (Why?)
Induction is a bit overkill here.
If $a \not = 1$,
$$\sum_{j=0}^n a^j=\frac{a^{n+1}-1}{a-1}$$
Here $a=-\frac 1 2$.
$$\frac{a^{n+1}-1}{a-1}=\frac{(-1/2)^{n+1}-1}{-1/2-1}=\frac{2}{3}\left(-\left(-\frac 1 2\right)^{n+1}+1\right)$$ $$=\frac{(-1)^n+2^{n+1}}{3 \cdot 2^n}$$
