Question

What is wrong with this “proof”? “Theorem” For every positive integer n, $\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathbf{ }\mathit{i}\mathbf{=}\frac{{\left(n+\frac{1}{2}\right)}^{\mathbf{2}}}{\mathbf{2}}\mathbf{.}$.

Basis Step: The formula is true for n = 1.

Inductive Step: Suppose that$\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathbf{ }\mathbf{i}\mathbf{=}\frac{{\left(n+\frac{1}{2}\right)}^{\mathbf{2}}}{\mathbf{2}}\mathbf{.}\mathbf{}\mathbf{Then}\mathbf{}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{ }\mathbf{i}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathbf{ }\mathbf{i}\mathbf{+}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}$. Then. By the inductive hypothesis,

$$\begin{gathered} \mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{n}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}}{\mathbf{2}}\mathbf{+}\mathbf{n}\mathbf{+}\mathbf{1} \\ \mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{n}\mathbf{+}\frac{\mathbf{9}}{\mathbf{4}}}{\mathbf{2}} \\ \mathbf{=}\frac{{\mathbf{(}\mathbf{n}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}{\mathbf{2}} \\ \mathbf{=}\frac{{\mathbf{\left(}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}} \end{gathered}$$

completing the inductive step.

Short answer

The basic step is wrong.

Step-by-step solution

Mathematical Induction

The principle of mathematical induction is to prove that$P\left(n\right)$ is true for all positive integer n in two steps.

1. Basic step: To verify that $P\left(1\right)$ is true.

2. Inductive step: To prove the conditional statement if $P\left(k\right)$is true then $P(k+1)$is true.

Wrong in the proof

Let $P\left(n\right)$be the statement $\sum _{i=1}^n i=\frac{{\left(n+\frac{1}{2}\right)}^2}{2}$.

Basic step:

Let$n=1$ in $P\left(n\right)$ then the value of LHS is $\sum _{i=1}^1 i=1$. Find the value of RHS as follows:

$$\begin{gathered} \begin{array}{r}\\ \\ \\ \end{array} \\ \frac{{\left(n+\frac{1}{2}\right)}^2}{2}=\frac{{\left(1+\frac{1}{2}\right)}^2}{2} \\ =\frac{{\left(\frac{3}{2}\right)}^2}{2} \\ =\frac{\left(\frac{9}{4}\right)}{2} \\ =\frac{9}{8} \end{gathered}$$

Here, the RHS value is $=\frac{9}{8}$which is not the same as the value of LHS.

Therefore, the basic step is wrong.