_10^n=1((528)/(n(n+1)(n+2))) is equal to:

Question

$\sum _{10}^{n=1}(\frac{528}{n(n+1)(n+2)})$ is equal to:

Accepted Answer

$S=\sum _{10}^{n=1}\frac{528}{n(n+1)(n+2)}$

We will simplify the general term using partial fractions.

Use the standard identity: $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$

So,

$\frac{528}{n(n+1)(n+2)}=528\cdot \frac{1}{2}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$

$=264(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$

Hence,

$S=264\sum _{10}^{n=1}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$

Write a few terms: $S=264[(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3})+(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4})+(\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5})+\cdots +(\frac{1}{10\cdot 11}-\frac{1}{11\cdot 12})]$

Now the middle terms cancel out telescopically.

So we get

$S=264(\frac{1}{1\cdot 2}-\frac{1}{11\cdot 12})$

$S=264(\frac{1}{2}-\frac{1}{132})$

Take LCM:

$\frac{1}{2}=\frac{66}{132}$

So,

$\frac{1}{2}-\frac{1}{132}=\frac{66-1}{132}=\frac{65}{132}$

Therefore,

$S=264\cdot \frac{65}{132}$

$S=2\cdot 65=130$

$130$

So, the correct option is Option B.

Solutions · 1