 # CAT Online Coaching – Partial Fractions

We get a lot of questions on partial fractions, on how to simplify them and how to solve questions that involve a summation of a series of fractional terms. In this post, I will solve 3 sums (in increasing order of difficulty) to explain the concept of partial fractions.

Try these questions on your own before you go to their solutions.

1. What is $\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +\frac { 1 }{ 4.5 } +...\frac { 1 }{ 99.100 }$

Solution: $\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +\frac { 1 }{ 4.5 } ...\frac { 1 }{ 99.100 } \\ \\ \frac { 1 }{ 1.2 } =\frac { 1 }{ 1 } -\frac { 1 }{ 2 } \\ \\ \frac { 1 }{ 2.3 } =\frac { 1 }{ 2 } -\frac { 1 }{ 3 } \\ \\ \frac { 1 }{ 99.100 } =\frac { 1 }{ 99 } -\frac { 1 }{ 100 }\\$

Therefore, the sum can be expressed as: $\left( \frac { 1 }{ 1 } -\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) +\left( \frac { 1 }{ 3 } -\frac { 1 }{ 4 } \right) ...\left( \frac { 1 }{ 99 } -\frac { 1 }{ 100 } \right) \\ \\ \frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 2 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } +...\frac { 1 }{ 99 } -\frac { 1 }{ 100 } \\ \\ \frac { 1 }{ 1 } -\frac { 1 }{ 100 } =\frac { 99 }{ 100 } \\ \\ \\ \\$

This question becomes easy once we spot that $\frac { 1 }{ 1.2 } =\frac { 1 }{ 1 } -\frac { 1 }{ 2 }$

How does one ‘spot’ this?

First step, write down the general nth term of the expression. In the above equation, ${ T }_{ n }=\frac { 1 }{ n(n+1) }$

Now, we start with the assumption that ${ T }_{ n }$ can be broken as two simpler fractions, say, like $\frac { 1 }{ n(n+1) } =\frac { A }{ n } +\frac { B }{ n+1 }$

Now, we need to find A and B. $\frac { A }{ n } +\frac { B }{ n+1 } =\frac { A(n+1)+B(n) }{ n(n+1) } =\frac { An+Bn+A }{ n(n+1) } =\frac { 1 }{ n(n+1) } \\$

Or, A + B = 0, A = 1. Or, A = 1, B = -1 $\frac { 1 }{ n(n+1) } =\frac { 1 }{ n } -\frac { 1 }{ n+1 } \\$

Now, try another question of a similar type:

2. Find $\frac { 1 }{ 3 } +\frac { 1 }{ 8 } +\frac { 1 }{ 15 } +\frac { 1 }{ 24 } +...\frac { 1 }{ 120 }$

Assuming that you have read and understood the concept behind the solution to the previous question, I am going to jump straight to the solution.

Each term can be expressed as: $\\ \frac { 1 }{ 1.3 } +\frac { 1 }{ 2.4 } +\frac { 1 }{ 3.5 } +\frac { 1 }{ 4.6 } +...\frac { 1 }{ 10.12 } \\ \\ \frac { 1 }{ 2 } \left( \frac { 1 }{ 1 } -\frac { 1 }{ 3 } \right) +\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } \right) +\frac { 1 }{ 2 } \left( \frac { 1 }{ 3 } -\frac { 1 }{ 5 } \right) +...\frac { 1 }{ 2 } \left( \frac { 1 }{ 10 } -\frac { 1 }{ 12 } \right) \\ \\ \frac { 1 }{ 2 } \left( \frac { 1 }{ 1 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } -\frac { 1 }{ 4 } +\frac { 1 }{ 3 } -\frac { 1 }{ 5 } +...\frac { 1 }{ 10 } -\frac { 1 }{ 12 } \right)$

After cancelling out the terms that appear as positive-negative pairs, we are left with: $\frac { 1 }{ 2 } \left( \frac { 1 }{ 1 } +\frac { 1 }{ 2 } -\frac { 1 }{ 11 } -\frac { 1 }{ 12 } \right) \\ \\ \frac { 1 }{ 2 } \left( \frac { 132+66-12-11 }{ 132 } \right) \\ \\ \frac { 1 }{ 2 } \left( \frac { 174 }{ 132 } \right) \\ \\ \frac { 87 }{ 132 }$

Now, try another question of a similar type but with a different way of solving.

3. If A = $\frac { 1 }{ 31.60 } +\frac { 1 }{ 32.59 } +\frac { 1 }{ 33.58 } ...+\frac { 1 }{ 59.32 } +\frac { 1 }{ 60.31 }$ and B = $\frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } +...+\frac { 1 }{ 59 } -\frac { 1 }{ 60 }$ find the value of $\frac { A }{ B }$.

This must be approached in a different way. Notice that in A, the sum of the terms in the denominator is equal. So, the terms can be written as: $\frac { 1 }{ 91 } \left( \frac { 1 }{ 31 } +\frac { 1 }{ 60 } \right) +\frac { 1 }{ 91 } \left( \frac { 1 }{ 32 } +\frac { 1 }{ 59 } \right) +...\frac { 1 }{ 91 } \left( \frac { 1 }{ 60 } +\frac { 1 }{ 31 } \right) \\ \\ \frac { 1 }{ 91 } \left( \frac { 1 }{ 31 } +\frac { 1 }{ 60 } +\frac { 1 }{ 32 } +\frac { 1 }{ 59 } +...\frac { 1 }{ 60 } +\frac { 1 }{ 31 } \right) \\ \\ \frac { 1 }{ 91 } \left( \frac { 2 }{ 31 } +\frac { 2 }{ 32 } +...\frac { 2 }{ 59 } +\frac { 2 }{ 60 } \right) \\ \\ \frac { 2 }{ 91 } \left( \frac { 1 }{ 31 } +\frac { 1 }{ 32 } +...\frac { 1 }{ 59 } +\frac { 1 }{ 60 } \right)$

We need to find a way to simplify B as well. Since there are some negative terms, let us try to eliminate them by adding and subtracting $\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 6 } +...+\frac { 1 }{ 60 }$

=> $\\ \frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } +...+\frac { 1 }{ 59 } -\frac { 1 }{ 60 } +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 6 } +...+\frac { 1 }{ 60 } \right) -\left( \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 6 } +...+\frac { 1 }{ 60 } \right) \\ \\ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } +...\frac { 1 }{ 60 } -2\left( \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 6 } +...+\frac { 1 }{ 60 } \right) \\ \\ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } +...\frac { 1 }{ 60 } -\left( \frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +...+\frac { 1 }{ 30 } \right) \\ \\ \frac { 1 }{ 31 } +\frac { 1 }{ 32 } +\frac { 1 }{ 33 } +...\frac { 1 }{ 60 }$

=> $\frac { A }{ B } =\frac { \frac { 2 }{ 91 } \left( \frac { 1 }{ 31 } +\frac { 1 }{ 32 } +...\frac { 1 }{ 59 } +\frac { 1 }{ 60 } \right) }{ \left( \frac { 1 }{ 31 } +\frac { 1 }{ 32 } +\frac { 1 }{ 33 } +...\frac { 1 }{ 60 } \right) } =\frac { 2 }{ 91 }$