Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $ Calculate $S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. This sequence is neither arithmetic nor geometric. How can you solve this. Thanks!
 A: Since $$\frac{1}{k(k+2)} = \frac{1}{2} \frac{2}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$$
for all $k$, we have
$$\sum_{k = 1}^n \frac{1}{k(k+1)(k+2)} = \frac{1}{2}\sum_{k = 1}^n \left(\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right),$$
which telescopes to $$\frac{1}{2}\left(\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right) = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}.$$
A: The method of partial fractions works for these. You have
$${1 \over x(x+1)(x +2)} = {1 \over 2} \bigg({1 \over x}\bigg) - {1 \over x + 1} + {1 \over 2} \bigg({1 \over x + 2}\bigg)$$
So your sum is the same as
$$\sum_{k = 1}^{n} \bigg[{1 \over 2} \bigg({1 \over k}\bigg) - {1 \over k + 1} + {1 \over 2} \bigg({1 \over k + 2}\bigg)\bigg]$$
This can be rewritten as 
$$\sum_{k = 1}^{n} \bigg[{1 \over 2}\bigg({1 \over k} - {1 \over k + 1}\bigg) + {1 \over 2}\bigg({1 \over k + 2} - {1 \over k + 1}\bigg)\bigg]$$
Now you have two telescoping sums which you can add. The result will be
$$\bigg({1 \over 2} - {1 \over 2(n + 1)}\bigg) + \bigg({1 \over 2(n + 2)} - {1 \over 4}\bigg)$$
$$= {1 \over 4} -{1 \over 2(n+1)(n+2)}$$
A: $$u_k = \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2k(k+1)} - \dfrac{1}{2(k+1)(k+2)} = 
f_k - f_{k+1}$$
so $$u_1 + u_2 + \cdots u_n = (f_1 - f_2) + (f_2 - f_3) + \cdots + (f_n - f_{n+1}) = f_1 - f_{n+1} = 
\dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)} 
$$
A: hint: use that $\frac{1}{k(k+1)(k+2)}=- \left( k+1 \right) ^{-1}+1/2\,{k}^{-1}+1/2\, \left( k+2 \right) ^{-1
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