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Rachita Aggarwal
$1+(1+a)r+(1+a+a^2)r^2+\cdots$ to $n$ terms.
$1+(1+a)r+(1+a+a^2)r^2+\cdots\\ =\dfrac{1}{1-a}\left[(1-a)+(1-a^2)r+(1-a^3)r^2+\cdots\right]\\ =\dfrac{1}{1-a}\left[(1+r+r^2+\cdots)-(a+a^2r+a^3r^2+\cdots)\right]\\ =\dfrac{1}{1-a}\left[\dfrac{1-r^n}{1-r}-\dfrac{a(1-a^nr^n)}{1-ar}\right]\\ (\text{where }a\ne1; r\ne 1; ar\ne 1)$