Question

Easy but cute question ,show that:-

\(\sum _{k=0}^n 10^k=\frac{10^{n+1}-1}{9}\)

Answer 1

:P

Answer 2

Induction: For \(n=0\), you have
\[\begin{align*}10^0=1&=\frac{10^1-1}{9}\\
&\stackrel{\checkmark}{=}1
\end{align*}\]
Assume
\[\sum_{k=0}^n10^k=\frac{10^{n+1}-1}{9}\]
Then
\[\begin{align*}\sum_{k=0}^{n+1}10^k&=\sum_{k=0}^{n}10^k+10^{n+1}\\\\
&=\frac{10^{n+1}-1}{9}+\frac{9\left(10^{n+1}\right)}{9}\\\\
&=\frac{10\left(10^{n+1}\right)-1}{9}\\\\
&=\frac{10^{n+2}-1}{9}\end{align*}\]
and we're done :)

Answer 3

nice