Prove the following summation-implication

Question

Can someone please show me to to prove this summation-implication?

$$\forall x \in \mathbb{Z^+}, x \neq 1 \implies \left(\forall n \in \mathbb{N}, \sum_{i=0}^{n-1}{x^i}=\frac{x^n-1}{x-1}\right) $$

Note: This is cleary a different one from Proving the geometric sum formula by induction so please stop down-voting.

Why don't you try induction on $n$? The base case is easy to check, and the inductive case is just addition of polynomials. — Sarvesh Ravichandran Iyer, Mar 06 '17 at 08:43
@астонвіллаолофмэллбэрг I don't see how that works. — Yahya Farooq, Mar 06 '17 at 08:44
Possible duplicate of http://math.stackexchange.com/questions/658992/proving-the-geometric-sum-formula-by-induction or http://math.stackexchange.com/questions/1945393/proving-the-formula-of-the-geometric-sum-through-induction. — Martin R, Mar 06 '17 at 08:45
@YahyaFarooq Can you at least see the base case? Take $n=1$, then the left hand side is $\sum_{i=0}^0 x^i = x^0=1$ while the right hand side is just $\frac{x-1}{x-1} = 1$. Similarly, assume the result for $n=k$ and prove it for $n=k+1$. All you need to do is to add $x^{k+1}$ to both sides then. — Sarvesh Ravichandran Iyer, Mar 06 '17 at 08:48
Possible duplicate of Proving the geometric sum formula by induction — Matthew Towers, Mar 06 '17 at 09:04
You have asked 4 questions in 3 hours. Don't you think you should slow down and take time to work by yourself ? — Jean Marie, Mar 06 '17 at 11:56

score 1 · Answer 1 · answered Mar 06 '17 at 09:31

1

Using identity $$x^n-y^n = (x-y)\cdot (\sum_{i=0}^{n-1}(x^i\cdot y^{n-1-i})$$ Putting y=1 $$\implies x^n-1 = (x-1)\cdot (\sum_{i=0}^{n-1}x^i \implies \sum_{i=0}^{n-1}{x^i}=\frac{x^n-1}{x-1}$$ Hence proved.

answered Mar 06 '17 at 09:31

sgrmshrsm7

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Prove the following summation-implication

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