Remainder when divided by $7$ What would be the remainder when
$12^1 +  12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.

As $12\equiv-2\pmod7, 12^3\equiv(-2)^3\equiv-8\equiv-1,$ord$_712=6$
$\implies12^{6k+r}\equiv12^r\equiv(-2)^r\pmod7$
$$\sum_{r=1}^{100}12^r\equiv12^1+12^2+12^3+12^4+16\sum_{r=0}^512^r\pmod7$$
$$\equiv(-2)+(-2)^2+(-2)^3+(-2)^4+2\sum_{r=0}^5(-2)^r\pmod7$$
Finally $\displaystyle\sum_{r=0}^5(-2)^r\equiv\dfrac{(-2)^6-1}{-2-1}\equiv0\pmod7$ as $(-2-1,7)=1$
