0

$$x^{\left(n-1\right)}+x^{\left(n-2\right)}c+x^{\left(n-3\right)}c^{2}+...+c^{\left(n-2\right)}+c^{\left(n-1\right)}=\frac{x^{n}-c^{n}}{x-c}$$ Where n is a positive integer

So step one is to test when $n=1$. I cannot even do this step. The $RHS = 1$ but I am struggling to get the LHS to also be one. I tried factorizing out $2$ after substituting since the end $x$ and $c$ become $1$, but I'm this doesn't really work. Is this a geometric progression?

Also is writing it out in this summation notation correct (I'm not too familiar with this notation, hopefully its right):

$$\sum _{n=n-1}^0\:x^n\:\cdot \:\sum _{n=0}^{n-1}c^n\:$$

user71207
  • 1,543
  • The LHS is also just $1$ when $n=1$. Can you see it? All $x$ powers and $c$ powers on the LHS sum up to $n-1$. For example, when $n=1$, LHS = $1$. When $n=2$, LHS = $x^1 c^0 + c^1 x^0$. – koifish Nov 10 '20 at 05:26
  • 1
    By the way, do you have to use induction? Because simply multiplying $(x-c)$ to the LHS and expansion yields the answer faster. – koifish Nov 10 '20 at 05:27
  • Yes but I dont understand why LHS = 1 when n = 1. $x^{1-1} = x^0 =1 $ and $c^{1-1} = c^0=1$ which gives 2 already? – user71207 Nov 10 '20 at 05:35
  • Also what do you mean by all the LHS powers sum up to $n-1$ – user71207 Nov 10 '20 at 05:52
  • 1
    Your summation is not correct. Partially because your indices don't make sense; 1: you can't put $n = n-1$, since this is impossible; 2: your dummy index should go from smaller numbers to larger, so doesn't make sense to go from $n-1$ (positive I assume) to 0. Also the formula is wrong, you should have a single summation. – xxxxxxxxx Nov 10 '20 at 06:21
  • Ahh ok oh well I won't use summation notation then – user71207 Nov 10 '20 at 06:24
  • Is it even possible to express it as a sum? – user71207 Nov 10 '20 at 06:24

1 Answers1

1

The base case $n=1$ is obviously true. Now suppose that it is true for some $r\in\Bbb{Z}^+,r\ge1$. The LHS can be written as $$\sum_{k=1}^{n}x^{n-k}c^{k-1}$$ We made the assumption that $$\sum_{k=1}^{r}x^{r-k}c^{k-1}=\frac{x^r-c^r}{x-c}$$ and we have to prove that $$\sum_{k=1}^{r+1}x^{r-k+1}c^{k-1}=\frac{x^{r+1}-c^{r+1}}{x-c}$$ Now \begin{align} \sum_{k=1}^{r+1}x^{r-k+1}c^{k-1}&=c^r+\sum_{k=1}^{r}x^{r-k+1}c^{k-1}\\ &=c^r+x\sum_{k=1}^{r}x^{r-k}c^{k-1} \end{align} we had made the assumption that $$\sum_{k=1}^{r}x^{r-k}c^{k-1}=\frac{x^r-c^r}{x-c}$$ So \begin{align} c^r+x\sum_{k=1}^{r}x^{r-k}c^{k-1}&=c^r+x\frac{x^r-c^r}{x-c}\\ &=c^r+\frac{x^{r+1}-xc^r}{x-c}\\ &=\frac{xc^r-c^{r+1}+x^{r+1}-xc^r}{x-c}\\ &=\frac{x^{r+1}-c^{r+1}}{x-c} \end{align} And hence the statement is true for $r+1$ if the statement is true for $r$. So by the principle of induction, the statement is true.