The sum, written in sigma notation, is ∞∑k=12k(k+1)(k+2).
We can perform partial fractions to get a telescoping sum.
Note that ∞∑k=1(f(k)−f(k+1))=f(1)−limk→∞f(k) for any function f.
Let 2k(k+1)(k+2)=Ak+Bk+1+Ck+2 for some constants A, B, C.
Multiplying both sides by k(k + 1)(k + 2) gives 2=A(k+1)(k+2)+Bk(k+2)+Ck(k+1).
Substituting k = 0, k = -1, and k = -2 respectively gives {2A=2−B=22C=2.
Then A = C = 1, B = -2.
We have:
∞∑k=12k(k+1)(k+2)=∞∑k=1(1k+1k+2−2k+1)=∞∑k=1(1k−1k+1)−∞∑k=1(1k+1−1k+2)
Can you continue with my hints here?