Why $\sum_{k=1}^n (\frac56)^{k-1}\cdot (\frac16)=1-(\frac56)^n$

Question

Why $$\sum_{k=1}^n \left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=1-\left(\frac56\right)^n$$??
Which formula I need to use to calculate it?

Thank you!

Answer 1

$$\sum_{k=1}^n\left(\frac56\right)^{k-1}\frac16=\frac15\sum_{k=1}^n\left(\frac56\right)^k\stackrel{\text{geom. series}}=\frac15\frac{\frac56-\left(\frac56\right)^{n+1}}{1-\frac56}=$$

$${}$$

$$=\frac65\left[\frac56-\left(\frac56\right)^{n+1}\right]=1-\left(\frac56\right)^n$$

Answer 2

Hint: The relevant formula is that of the geometric sum: $$\sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}$$

Answer 3

$\left(1-x\right)\sum_{k=1}^{n}x^{k-1}=\sum_{k=1}^{n}x^{k-1}-\sum_{k=1}^{n}x^{k}=\left(1+\dots+x^{n-1}\right)-\left(x^{1}+\dots+x^{n}\right)=1-x^{n}$.

Substitute $x=\frac{5}{6}$.