Okay I know I have the answer incorrect but not sure why...
Calculate the following series:
$$\sum_{j=0}^n 2^j \text{ for } n = 1,2,3,4$$
so I think it would be $2^1 = 2$ but the answer is $3$ and the rest are $7,9,32$... I do not understand how they got $3$. Any guidance will help.
The geometric series in general is $$\sum_{j=0}^n a^j = \frac{1-a^{n+1}}{1-a}.$$
For $a=2$, you get $$\sum_{j=0}^n 2^j = 2^{n+1}-1.$$
If your sum begins at $j=1$, $$\sum_{j=1}^n a^j = \frac{1-a^{n+1}}{1-a}-1=\frac{a-a^{n+1}}{1-a}.$$ If you use $a=2$, you get $$\sum_{j=1}^n 2^j =2^{n+1}-2.$$
$$ \sum_{j=0}^12^j=2^0+2^1=3 $$ $$ \sum_{j=0}^22^j=2^0+2^1+2^2=7 $$ $$ \sum_{j=0}^32^j=2^0+2^1+2^2+2^3=15 \qquad[\text{not }9] $$ $$ \sum_{j=0}^42^j=2^0+2^1+2^2+2^3+2^4=31 \qquad [\text{not }32] $$
