Evaluating the Sum of Fractions

We are asked to evaluate the sum of the following fractions:

\(\frac 1 {15} + \frac 1 {35} + \frac 1 {63} + \frac 1 {99} + \frac 1 {143}\)

Identifying the Pattern in Denominators

Let's examine the denominators:

• \(15 = 3 \times 5\)
• \(35 = 5 \times 7\)
• \(63 = 7 \times 9\)
• \(99 = 9 \times 11\)
• \(143 = 11 \times 13\)

We observe that each denominator is a product of two consecutive odd numbers. The fractions can be written in the form \(\frac{1}{n(n+2)}\), where \(n\) is an odd number starting from 3.

Using Partial Fraction Decomposition for Sum Evaluation

A fraction of the form \(\frac{1}{n(n+2)}\) can be decomposed into partial fractions. We look for constants A and B such that:

\(\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}\)

Multiplying by \(n(n+2)\), we get:

\(1 = A(n+2) + Bn\)

Setting \(n=0\), we get \(1 = A(0+2) + B(0) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}\).

Setting \(n=-2\), we get \(1 = A(-2+2) + B(-2) \Rightarrow 1 = -2B \Rightarrow B = -\frac{1}{2}\).

So, the decomposition is:

\(\frac{1}{n(n+2)} = \frac{1/2}{n} + \frac{-1/2}{n+2} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right)\)

Applying Decomposition to Each Term

Let's apply this decomposition to each term in the sum:

• \(\frac{1}{15} = \frac{1}{3 \times 5} = \frac{1}{2} \left(\frac{1}{3} - \frac{1}{5}\right)\)
• \(\frac{1}{35} = \frac{1}{5 \times 7} = \frac{1}{2} \left(\frac{1}{5} - \frac{1}{7}\right)\)
• \(\frac{1}{63} = \frac{1}{7 \times 9} = \frac{1}{2} \left(\frac{1}{7} - \frac{1}{9}\right)\)
• \(\frac{1}{99} = \frac{1}{9 \times 11} = \frac{1}{2} \left(\frac{1}{9} - \frac{1}{11}\right)\)
• \(\frac{1}{143} = \frac{1}{11 \times 13} = \frac{1}{2} \left(\frac{1}{11} - \frac{1}{13}\right)\)

Summing the Decomposed Terms (Telescoping Sum)

Now, let's write the sum using the decomposed terms:

\(\text{Sum} = \frac{1}{2} \left(\frac{1}{3} - \frac{1}{5}\right) + \frac{1}{2} \left(\frac{1}{5} - \frac{1}{7}\right) + \frac{1}{2} \left(\frac{1}{7} - \frac{1}{9}\right) + \frac{1}{2} \left(\frac{1}{9} - \frac{1}{11}\right) + \frac{1}{2} \left(\frac{1}{11} - \frac{1}{13}\right)\)

We can factor out the \(\frac{1}{2}\):

\(\text{Sum} = \frac{1}{2} \left[\left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{9}\right) + \left(\frac{1}{9} - \frac{1}{11}\right) + \left(\frac{1}{11} - \frac{1}{13}\right)\right]\)

Notice that the intermediate terms cancel each other out:

\(\text{Sum} = \frac{1}{2} \left[\frac{1}{3} \cancel{- \frac{1}{5}} \cancel{+ \frac{1}{5}} \cancel{- \frac{1}{7}} \cancel{+ \frac{1}{7}} \cancel{- \frac{1}{9}} \cancel{+ \frac{1}{9}} \cancel{- \frac{1}{11}} \cancel{+ \frac{1}{11}} - \frac{1}{13}\right]\)

\(\text{Sum} = \frac{1}{2} \left[\frac{1}{3} - \frac{1}{13}\right]\)

This is known as a telescoping sum.

Calculating the Final Value

Now, we calculate the difference inside the brackets:

\(\frac{1}{3} - \frac{1}{13} = \frac{1 \times 13 - 1 \times 3}{3 \times 13} = \frac{13 - 3}{39} = \frac{10}{39}\)

Substitute this back into the sum expression:

\(\text{Sum} = \frac{1}{2} \times \frac{10}{39}\)

\(\text{Sum} = \frac{10}{2 \times 39} = \frac{10}{78}\)

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:

\(\text{Sum} = \frac{10 \div 2}{78 \div 2} = \frac{5}{39}\)

Thus, the value of the given sum is \(\frac{5}{39}\).

Revision Table: Key Concepts

Additional Information: Extending the Sum of Fractions

The pattern observed in this sum is common in mathematical series. If the sum were extended to more terms following the same pattern, for example:

\(\frac 1 {3 \times 5} + \frac 1 {5 \times 7} + \frac 1 {7 \times 9} + \dots + \frac{1}{(2k+1)(2k+3)}\)

Using the partial fraction decomposition \(\frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right)\), the sum of \(m\) terms would be:

\(\frac{1}{2} \left[\left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots + \left(\frac{1}{2m+1} - \frac{1}{2m+3}\right)\right]\)

The intermediate terms cancel out, leaving:

\(\frac{1}{2} \left(\frac{1}{3} - \frac{1}{2m+3}\right)\)

In our specific problem, the terms are for \(n=3, 5, 7, 9, 11\). The last term is \(\frac{1}{11 \times 13}\), where the first number in the denominator is 11. If we map this to \(2m+1\), \(2m+1 = 11 \Rightarrow 2m = 10 \Rightarrow m = 5\). So there are 5 terms in the sum. Using the general formula with \(m=5\):

\(\frac{1}{2} \left(\frac{1}{3} - \frac{1}{2(5)+3}\right) = \frac{1}{2} \left(\frac{1}{3} - \frac{1}{13}\right) = \frac{1}{2} \left(\frac{13-3}{39}\right) = \frac{1}{2} \times \frac{10}{39} = \frac{5}{39}\)

This confirms our step-by-step calculation and shows how the pattern works for any number of terms following this structure.