Hey want a detailed hint about how to solve this question

Question

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

Answer 1

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.

Answer 2

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 $1$.