0

I was curious about the formula for the difference between a number powered by a certain number and the sum of the number powered from 0 to the number - 1. I found this formula but I'm not sure if this is correct, and if it was correct, I couldn't prove it in a "proper" way. Can someone here show me the proof of this formula because I'm really curious about it?

$$x^n = \left[(x-1)\sum\limits_{k=0}^{n-1}x^k\right]+1$$

vitamin d
  • 5,783
Edward
  • 3
  • You are asking why $99999+1 = 100000$? – JMoravitz Aug 13 '21 at 14:46
  • 2
    See https://en.wikipedia.org/wiki/Geometric_progression#Related_formulas – Robert Z Aug 13 '21 at 14:48
  • The sum $(x-1)\sum\limits_{k=0}^{n-1}x^k = \sum\limits_{k=0}^{n-1}(x^{k+1} - x^k)$ is a telescoping sum. A lot of cancellations appear among the terms and you can read off the end result by replacing the $k$ in $x^{k+1}$ by the largest possible $k$ and the $k$ in $x^k$ by the smallest possible $k$. This gives you $(x-1)\sum\limits_{k=0}^{n-1}x^k = x^{(n-1)+1} - x^0 = x^n - 1$. Look at above wiki entries for more examples of this sort of sums. – achille hui Aug 13 '21 at 16:04

2 Answers2

1

Given a finite series $$S=\sum_{k=0}^{n-1} x^k=1+x+x^2+\ldots+x^{n-1}$$ We want to find a closed expression for $S$. Note that by multiplying by $x$, we get $$Sx=x+x^2+x^3+\ldots+x^n$$ Now subtract both eqations and note the massive cancellation, $$(x-1)S=x^n-1$$ $$(x-1)S+1=x^n$$ $$\left[(x-1)\sum_{k=0}^{n-1} x^k\right]+1=x^n$$

Alan Abraham
  • 5,142
  • 6
  • 20
0

Do you know the formula to factor a polynomial in the form of $a^n - b^n$ ?

$x^n - 1^n = x^n - 1 = (x-1)(\sum _{k=0}^{n-1}\:x^k1^{n-1-k}) = (x-1)(\sum _{k=0}^{n-1}\:x^k)$

You take the expression above and you add $+ 1$ to get $x^n$

sky1099
  • 91