Show that : $1 + ( 1 - \frac{x}{n}) + ...... + ( 1 - \frac{x}{n})^{n-1} = \frac{n}{x}(1-(1-\frac{x}{n})^{n}) $

Question

How can I prove that : $\,\forall\,n \ge1 \, ,x>0 $

$$1 + ( 1 - \frac{x}{n}) + ...... + ( 1 - \frac{x}{n})^{n-1} = \frac{n}{x}(1-(1-\frac{x}{n})^{n}) \;\;\;\:\:\:$$

thanks in advance .

Answer 1

you try to calculate $$\sum_{k=0}^{n-1}{p}^k$$ with $p=1-\frac{x}{n}$ We know that $$\sum_{k=0}^{n-1}{p}^k=\frac{1-p^{n}}{1-p}$$

Therefore, $$\sum_{k=0}^{n-1}{p}^k=\frac{1-\left(1-\frac{x}{n}\right)^{n}}{1-(1-\frac{x}{n})}=\frac{n}{x}(1-(1-\frac{x}{n})^{n})$$

Answer 2

It is a geometric series \begin{eqnarray*} \sum_{i=0}^{n-1} X^i = \frac{1-X^n}{1-X}. \end{eqnarray*} Let $X=(1-x/n)$ and we have \begin{eqnarray*} 1 + \left( 1 - \frac{x}{n} \right) + ...... + \left( 1 - \frac{x}{n}\right)^{n-1} = \frac{1-(1-\frac{x}{n})^n}{1-(1-\frac{x}{n})} \\ = \frac{n}{x} \left(1-\left(1-\frac{x}{n}\right)^{n}\right) \end{eqnarray*}

Answer 3

In general, the following formula is true (which you can confirm through mathematical induction):$\sum_{i=0}^k X^i=\frac{1-X^{k+1}}{1-X}.$

Substituting $X=1-\frac xn$ and $k=n-1,$ we obtain:

\begin{equation} \begin{split} 1+(1-\frac xn)+\cdots+(1-\frac xn)^{n-1}&=\frac{1-(1-\frac xn)^n}{1-(1-\frac xn)}\\ &=\frac nx\left(1-{\left(1-\frac xn\right)}^n\right).\\ \end{split} \end{equation}

Answer 4

From the high school formula: \begin{align} &&1-u^n&=(1-u)(1+u+u^2+\dots+u^{n-1})\\[1ex] &\text{whence }\qquad& 1+u+u^2+\dots+u^{n-1}&=\frac{1-u^n}{1-u}. \end{align} Now set $\;u=1-\dfrac xn$ to obtain $$1 + \Bigl( 1 - \frac{x}{n}\Bigr) + … + \Bigl( 1 - \frac{x}{n}\Bigr)^{n-1}=\frac{1-\Bigl(1-\cfrac{x}{n}\Bigr)^{n}}{\not1-\Bigl(\not1-\cfrac xn\Bigr)}=\frac nx\biggl(1-\Bigl(1-\cfrac{x}{n}\Bigr)^{n}\biggr).$$