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**mark1950** Find the value of $\displaystyle \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) = $\displaystyle -\frac{1}{r + 2}$

f(r - 1) = $\displaystyle -\frac{1}{r+1} $

Hence,

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

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

Where have I gone wrong?