# Thread: [SOLVED] How do I find this value?

1. ## [SOLVED] How do I find this value?

Find the value of $\frac{1}{2.3} + \frac{1}{3.4} + \frac {1}{4.5} + ... + \frac{1}{50.51}$

THis is my take on the question:
n = 49

Using method of difference,

f(r) = $-\frac{1}{r + 2}$
f(r - 1) = $-\frac{1}{r+1}$
Hence,
f(n) - f(1 - 1) = $1 - \frac{1}{n + 2}$

When I sub n = 49 into this equation, I found $\frac{50}{51}$ which is not the same as the answer of $\frac{49}{102}$

Where have I gone wrong?

2. Originally Posted by mark1950
Find the value of $\frac{1}{2.3} + \frac{1}{3.4} + \frac {1}{4.5} + ... + \frac{1}{50.51}$

THis is my take on the question:
n = 49

Using method of difference,

f(r) = $-\frac{1}{r + 2}$
f(r - 1) = $-\frac{1}{r+1}$
Hence,
f(n) - f(1 - 1) = $1 - \frac{1}{n + 2}$

When I sub n = 49 into this equation, I found $\frac{50}{51}$ which is not the same as the answer of $\frac{49}{102}$

Where have I gone wrong?
Hi

Firstly , we know the general term is $\frac{1}{(r+1)(r+2)}$

Then do partial fraction here , we get

$\sum^{49}_{r=1}[\frac{1}{r+1}-\frac{1}{r+2}]$

Here let $f(r)=\frac{1}{r+1}$ , then $f(r+1)=\frac{1}{r+2}$

so

$\sum^{49}_{r=1}[f(r)-f(r+1)]$

$= f(1)-f(50)$

$=\frac{1}{1+1}-\frac{1}{50+1}$

Continue from here , you should get the answer .