CAT Online Coaching – Partial Fractions

Number Theory Leave a comment

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 }  

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