Question

Is this proof that \(\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \cdots + \frac{1}{{\left( {n - 1} \right)n}} = \frac{3}{2} - \frac{1}{n}\),

whenever \(n\) is a positive integer, correct? Justify your answer.

Basis step: The result is true when \(n = 1\). Because

\(\frac{1}{{1 \cdot 2}} = \frac{3}{2} - \frac{1}{1}\).

Inductive step: Assume the result is true for \(n\). Then

\(\begin{aligned}{c}\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \cdots + \frac{1}{{\left( {n - 1} \right)n}} + \frac{1}{{n\left( {n + 1} \right)}} &= \frac{3}{2} - \frac{1}{n} + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)\\ &= \frac{3}{2} - \frac{1}{{n + 1}}\end{aligned}\)

Hence the result is true for \(n + 1\) if it is true for \(n\). This completes the proof.

Short answer

The basis step is incorrect.

Step-by-step solution

To recall the concepts and definition

Mathematical Induction:

The mathematical induction is defined as follows:

Step 1 (Base step): In this step, to prove that the statement is true for n=1.

Step 2(Inductive step): In this case, if the statement is true for n^th iteration, then to prove it is also true for (n+1)^st iteration.

To check whether the proof is correct or incorrect

The basis step is incorrect.

This is because the term \(\frac{1}{{1 \cdot 2}}\) is of the form \(\frac{1}{{\left( {n - 1} \right)n}}\) with \(n = 2\).

Thus, the fraction \(\frac{1}{{1 \cdot 2}}\) should not be used for the basis step with \(n = 1\).