Simplifying the Given Mathematical Series

The question asks us to simplify the following mathematical expression:

\({{1} \over 3\times4} + {{1} \over 4\times5} + {{1} \over 5\times6} + ... + ...............{{1} \over 20\times21}\)

This expression is a sum of terms where each term is of the form \({{1} \over n(n+1)}\). The series starts from \(n=3\) and goes up to \(n=20\).

We can write the general term \({{1} \over n(n+1)}\) using partial fraction decomposition. The idea is to express this fraction as the difference of two simpler fractions:

\({{1} \over n(n+1)} = {A \over n} + {B \over n+1}\)

To find the values of \(A\) and \(B\), we can combine the fractions on the right side:

\({{1} \over n(n+1)} = {A(n+1) + Bn \over n(n+1)}\)

Comparing the numerators, we get:

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

\(1 = An + A + Bn\)

\(1 = (A+B)n + A\)

For this equation to hold for all values of \(n\), the coefficients of \(n\) on both sides must be equal, and the constant terms must be equal. This gives us a system of linear equations:

1. \(A+B = 0\)
2. \(A = 1\)

From equation (2), we have \(A=1\). Substituting this into equation (1), we get \(1+B=0\), which means \(B = -1\).

So, the partial fraction decomposition is:

\({{1} \over n(n+1)} = {1 \over n} - {1 \over n+1}\)

Now we can apply this decomposition to each term in the given series:

• For the first term (\(n=3\)): \({{1} \over 3\times4} = {{1} \over 3} - {{1} \over 4}\)
• For the second term (\(n=4\)): \({{1} \over 4\times5} = {{1} \over 4} - {{1} \over 5}\)
• For the third term (\(n=5\)): \({{1} \over 5\times6} = {{1} \over 5} - {{1} \over 6}\)
• ...
• For the last term (\(n=20\)): \({{1} \over 20\times21} = {{1} \over 20} - {{1} \over 21}\)

Let's write out the sum using these decomposed terms:

\(\left({{1} \over 3} - {{1} \over 4}\right) + \left({{1} \over 4} - {{1} \over 5}\right) + \left({{1} \over 5} - {{1} \over 6}\right) + ... + \left({{1} \over 20} - {{1} \over 21}\right)\)

Notice that the second part of each term cancels out the first part of the next term. This type of series is called a telescoping series. The terms \(-{{1} \over 4}\) and \(+{{1} \over 4}\) cancel, \(-{{1} \over 5}\) and \(+{{1} \over 5}\) cancel, and so on, until \(-{{1} \over 20}\) and \(+{{1} \over 20}\) cancel.

The only terms that remain are the very first part of the first term and the very last part of the last term:

\({{1} \over 3} - {{1} \over 21}\)

Now, we just need to subtract these fractions. To do this, we find a common denominator, which is 21.

\({{1} \over 3} = {{1 \times 7} \over {3 \times 7}} = {{7} \over 21}\)

So the expression becomes:

\({{7} \over 21} - {{1} \over 21}\)

\(= {{7-1} \over 21}\)

\(= {{6} \over 21}\)

Finally, we simplify the fraction \({{6} \over 21}\) by dividing both the numerator and the denominator by their greatest common divisor, which is 3.

\({{6 \div 3} \over {21 \div 3}} = {{2} \over 7}\)

Therefore, the result of simplifying the given expression is \({{2} \over 7}\).

Revision Table: Key Concepts

Additional Information: Understanding Series and Partial Fractions

What is a Series?

In mathematics, a series is the sum of the terms of a sequence. For example, if you have a sequence \(a_1, a_2, a_3, ...\), the corresponding series is \(a_1 + a_2 + a_3 + ...\). Our problem deals with a finite series, meaning it has a limited number of terms, starting from \(n=3\) and ending at \(n=20\).

Partial Fraction Decomposition Explained

This technique is used to integrate rational functions in calculus, but it's also very useful in simplifying sums of series like the one in this problem. The core idea is that a fraction like \({{1} \over (x-a)(x-b)}\) can be written as a sum or difference of simpler fractions like \({{A} \over x-a} + {{B} \over x-b}\). For the specific form \({{1} \over n(n+1)}\), the decomposition is always \({{1} \over n} - {{1} \over n+1}\), which is a very common pattern seen in telescoping series problems.

More on Telescoping Series

The term "telescoping" comes from the way a telescoping telescope collapses – the segments fit inside each other, and only the outer ends remain visible. In a telescoping series, when you write out the sum, the inner terms cancel each other out. For a general telescoping series \(\sum_{n=k}^{m} (f(n) - f(n+1))\), the sum is \(f(k) - f(m+1)\). In our case, \(f(n) = {{1} \over n}\), \(k=3\), and \(m=20\). The sum is \(f(3) - f(21) = {{1} \over 3} - {{1} \over 21}\), which matches our calculation.