How to find the sum of sequence $ 1+4+4^2+\cdots+4^{X+Y} $?

Question

I see the following sequence and it's: $$h=1+4+4^2+\cdots+4^{X+Y}=\frac{4^{X+Y+1}-1}{4-1}$$

how we get this sequence?

I know this is a primary question but I confused :)

Answer 1

This is an example of a geometric series. Let's say we try to sum: $$S=1+r+r^2+r^3+\dotsb+r^n$$ In your example, $r=4$, and $n=X+Y$.

There is a well known "trick" for solving this. Multiply by $r$: $$Sr=r+r^2+r^3+\dotsb+r^n+r^{n+1}$$ Notice how this is the same thing as $S$, except without the $1$ at front and with an extra $r^{n+1}$ at the end. In fact: $$Sr=S-1+r^{n+1}$$ Solving: \begin{align} Sr&=S-1+r^{n+1} \\ Sr-S&=-1+r^{n+1} \\ S(r-1)&=r^{n+1}-1 \\ S&=\frac{r^{n+1}-1}{r-1} \end{align}

$$\fbox{$1+r+r^2+r^3+\dotsb+r^n=\dfrac{r^{n+1}-1}{r-1}$}$$

Plugging in $r=4$ and $n=X+Y$, we get your answer.

Answer 2

$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$

Answer 3

This is a GP, where $a= $ first term=$1$, $r=$common ratio=$4$

Sum of GP is given by $\frac{a(r^n-1)}{r-1}$

$1+\underbrace{4+4^2+\cdots+4^{X+Y}}$
$1+\hspace{25 pt}X+Y\hspace{30 pt}$=n=Number of terms