My son started high-school, so I’m helping him with mathematics. It’s been a long time, and I have to re-learn. But it’s fun, and I’m hoping to start a series of articles.

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Here’s an interesting sum:

\[S = \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \cdots + \frac{1}{99 \cdot 103}\]

In order to write it with the summation notation, we first need to observe that the difference between elements in this number sequence is 4, therefore 4k is involved in the formula of the general term:

\[S = \sum_{k=1}^{25} \frac{1}{(4k - 1)(4k + 3)}\]

Such sums are actually telescopic. And when you have fractions like these with a constant numerator, the general term can be split in 2 fractions, with the help of an equation, needing to find an a satisfying this:

\[\frac{1}{(4k-1)(4k+3)} = a \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right)\] \[a \left( (4k+3) - (4k-1) \right) = a \cdot 4 = 1 \implies a = \frac{1}{4}\]

The sum becomes:

\[\begin{align*} S &= \sum_{k=1}^{25} \frac{1}{(4k-1)(4k+3)} \\ &= \sum_{k=1}^{25} \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right) \\ &= \frac{1}{4} \sum_{k=1}^{25} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right) \end{align*}\]

Expanding it:

\[S = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \frac{1}{11} - \frac{1}{15} + \ldots + \frac{1}{99} - \frac{1}{103} \right)\]

The terms get cancelled, the result being:

\[S = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{103} \right)\]

More samples #

We can apply this solution to other similar sums as well, for example:

\[\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)(2k+3)}\]

In which case the fraction decomposition can be found with this equation:

\[\frac{1}{(2k-1)(2k+1)(2k+3)} = a \left( \frac{1}{(2k-1)(2k+1)} - \frac{1}{(2k+1)(2k+3)} \right)\]

And it works with more denominator factors as well:

\[\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)(k+3)}\]

With the equation for finding the fraction decomposition being:

\[\frac{1}{k(k+1)(k+2)(k+3)} = a \left( \frac{1}{k(k+1)(k+2)} - \frac{1}{(k+1)(k+2)(k+3)} \right)\]

Non-constant numerators #

For this one we no longer have a constant numerator and the above solution no longer works:

\[S = \sum_{k=1}^n \frac{7k^2 + k}{(7k-3)(7k+4)}\]

To do this fraction decomposition, we now need to find 2 constants, a and b, the equation now being:

\[\frac{7k^2 + k}{(7k-3)(7k+4)} = a + \frac{b}{7k-3} - \frac{b}{7k+4}\]

To solve it, after eliminating the denominator, the trick is to group by powers of k:

\[\begin{align*} 7k^2 + k &= a(7k-3)(7k+4) + b(7k+4) - b(7k-3) \\ 7k^2 + k &= a(49k^2 + 28k - 21k - 12) + 7kb + 4b - 7kb + 3b \\ 0 &= 49a k^2 + 7a k + 7b - 12a \\ 0 &= k^2 (49a-7) + k(-1+7a) + (7b - 12a) \end{align*}\]

k being variable, it means that, in order for the above equation to have solutions, we need the constants to nullify k, so we have this system of equations:

\[\left\{ \begin{array}{l} 49a - 7 = 0 \\ 7a - 1 = 0 \\ 7b - 12a = 0 \end{array} \right.\] \[\begin{align*} a &= \frac{1}{7} \\ 7b - \frac{12}{7} &= 0 \\ 7b &= \frac{12}{7} \\ b &= \frac{12}{49} \end{align*}\]

And now we can finally write the sum as:

\[\begin{align*} S &= \sum_{k=1}^{n} \left( \frac{1}{7} + \frac{12}{49} \left( \frac{1}{7k-3} - \frac{1}{7k+4} \right) \right) \\ &= \frac{n}{7} + \frac{12}{49} \sum_{k=1}^{n} \left( \frac{1}{7k-3} - \frac{1}{7k+4} \right) \\ &= \frac{n}{7} + \frac{12}{49} \left( \frac{1}{4} - \frac{1}{7n+4} \right) \end{align*}\]

Published: October 15, 2025 | Written by Alexandru Nedelcu