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The question is: To prove the following summation for every positive integer n:

$$\sum_{k=0}^n x^k = ({n+1}){..if..x =1 }$$ and a similar one which is: $$\sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1} {..if..≠ 1}$$

image of question if u don't understand

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closed as off-topic by trancelocation, Ernie060, Alexander Gruber Apr 29 at 23:04

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    $\begingroup$ The first one (case $x = 1$) is obvious really. Have you tried plugging in $x=1$ into the LHS and writing out the sum? How many terms do you have, what is the value of each of them? For the second one, have you heard of induction? Alternatively, you can prove it using the proof for the sum of a geometric series. See en.wikipedia.org/wiki/Geometric_series#Formula $\endgroup$ – PhysicsMathsLove Apr 29 at 13:45
  • $\begingroup$ did first one and got the answer but for second one it's a bit confusing for me,as you said im searching for induction $\endgroup$ – AbdurRafay Apr 29 at 14:01
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    $\begingroup$ See Wikipedia on geometric series for a proof $\endgroup$ – Ross Millikan Apr 29 at 14:06
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As PhysicsMathsLove said, the first case is trivial, really. For you not to divide by zero in the second case, you have to make that distinction $x \neq 1$. The rest is proof by induction. First, we prove that (n=1) holds: $$\sum_{k=0}^{n}x^k = x^1 + x^0 = x + 1 = \frac{(x+1)(x-1)}{x-1} = \frac{x^2-1}{x-1}$$.

Now, we go from $n \rightarrow n+1$, assuming our hypothesis holds for $n$: $$\sum_{k=0}^{n+1} x^k = \sum_{k=0}^n x^k + x^{n+1} =_{IH} \frac{x^{n+1}-1}{x-1} + x^{n+1}$$ $$= \frac{x^{n+1}-1}{x-1} + \frac{x^{n+1}(x-1)}{x-1} = \frac{x^{n+1}-1 + x^{n+2}-x^{n+1}}{x-1} = \frac{x^{n+2}-1}{x-1}$$ q.e.d.

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  • $\begingroup$ what will happen if n=1 is undefined? $\endgroup$ – AbdurRafay Apr 29 at 15:10
  • $\begingroup$ Then the second case doesn't make any sense at all, which is why you wrote $...if x \neq 1$ in the first place. $\endgroup$ – TheEye Apr 29 at 15:15
  • $\begingroup$ thanks bro for helping me $\endgroup$ – AbdurRafay Apr 29 at 15:22
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Hint : \begin{align*} S &= \sum_{k=0}^n x^k \\ &= 1 - x^{n+1} + \sum_{k=1}^{n+1} x^k\\ &= 1 - x^{n+1} + x S\\ \end{align*} Solving for $S$ gives you the second equality. First equality is just replacing $x$ with $1$.

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