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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 :)

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  • $\begingroup$ Do you mean, how does one prove the formula for the sum of terms? $\endgroup$ Aug 24, 2014 at 16:55
  • $\begingroup$ Dear @Travis, I means how get this? $\endgroup$ Aug 24, 2014 at 16:56

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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.

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  • $\begingroup$ can we say sum is: $((4^{X+Y+1}-1)/3)-1?$ $\endgroup$ Aug 24, 2014 at 17:28
  • $\begingroup$ No. Why do you have the $-1$ at the end? It should just be: $$\frac{4^{X+Y+1}-1}{4-1}=\frac{4^{X+Y+1}-1}{3}$$without any $-1$s at the end. (If you were summing $4+4^2+\cdots+4^{X+Y}$ rather than $1+4+4^2+\cdots+4^{X+Y}$, then sure.) $\endgroup$ Aug 24, 2014 at 17:33
  • $\begingroup$ you means $4+...+4^X+Y=((4^{X+Y+1}-1)/3)-1 $ am I right? $\endgroup$ Aug 24, 2014 at 17:35
  • $\begingroup$ Yes. If we were to start with $4+\dotsb$ rather than $1+4+\dotsb$, then you would have to subtract off the $1$. However, in your Original Post, you asked the question with the $1$. $\endgroup$ Aug 24, 2014 at 17:37
  • $\begingroup$ yes, thankssss so muchhhhh $\endgroup$ Aug 24, 2014 at 17:38
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$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$

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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

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