Question

The unit digit of $\left(1+9+9^2+9^3+.....+9^{2006}\right)$ is.....

Answer 1

$9^1=9$

$9^2=81$

$9^3=729$

$9^4=6561$ and so on.......

$\to$ If we notice this pattern the unit place has $1$ and $9$ alternatively. So if we add consecutive like $9^1+9^2$ and $9^3+9^4$ we ll get $1003$ terms with $0$ as unit digits

$\to$ This will leave: $1+(1003$ terms with $0$ as unit digit )

$\Rightarrow$ Thus $1$ will be the digit at last place in the final term.