$$\sum_{r=2}^{100}\dfrac{3^r(2-2r)}{r(r+1)(r+2)}$$
My attempt is as follows:-
$$-2\sum_{r=2}^{100}\dfrac{3^r(r-1)}{r(r+1)(r+2)}$$
$$-2\sum_{r=2}^{100}\dfrac{3^r}{(r+1)(r+2)}-\dfrac{3^r}{r(r+1)(r+2)}$$
$$-2\sum_{r=2}^{100}\dfrac{3^r}{(r+1)(r+2)}-\dfrac{1}{2}\dfrac{3^r\cdot(r+2-r)}{r(r+1)(r+2)}$$
$$-2\sum_{r=2}^{100}\dfrac{3^r}{(r+1)(r+2)}-\dfrac{1}{2}\left(\dfrac{3^r}{r(r+1)}-\dfrac{3^r}{(r+1)(r+2)}\right)$$
$$-2\sum_{r=2}^{100}\dfrac{3}{2}\cdot\dfrac{3^r}{(r+1)(r+2)}-\dfrac{1}{2}\left(\dfrac{3^r}{r(r+1)}\right)$$
$$-\sum_{r=2}^{100}{3}\cdot\dfrac{3^r}{(r+1)(r+2)}-\left(\dfrac{3^r}{r(r+1)}\right)$$
$$-\sum_{r=2}^{100}{3}\cdot\left(\dfrac{3^r}{r+1}-\dfrac{3^r}{r+2}\right)-\left(\dfrac{3^r}{r}-\dfrac{3^r}{r+1}\right)$$
$$-\sum_{r=2}^{100}{4}\cdot\dfrac{3^r}{r+1}-3\cdot\dfrac{3^r}{r+2}-\dfrac{3^r}{r}$$
Now let's assume $S_1=\dfrac{3^r}{r+1}, S_2=\dfrac{3^r}{r+2}, S_3=\dfrac{3^r}{r}$
$$S_1=\dfrac{3^2}{3}+\dfrac{3^3}{4}+\dfrac{3^4}{5}\cdots\cdots$$
$$S_2=\dfrac{3^2}{4}+\dfrac{3^3}{5}+\dfrac{3^4}{6}\cdots\cdots$$
$$S_3=\dfrac{3^2}{2}+\dfrac{3^3}{3}+\dfrac{3^4}{4}\cdots\cdots$$
Now here I am not getting how to calculate $S_1,S_2,S_3$. Please help me in this.