Chapter 12.5, Problem 17E

To determine

To Prove: $n^2-n+41$ is Odd for all natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First, we need to prove the above formula is True for $n=1$ .

Given formula is

  $n^2-n+41$ is Odd for all natural numbers.

Put $n=1$

We get

  $\begin{array}{l}{n}^2-n+41=1^2-1+41\\ \Rightarrow n^2-n+41=1-1+41\\ \Rightarrow n^2-n+41=41\end{array}$

41 is an odd number

It True for $n=1$ .

Let us assume it is True for $n=k$

So

  $k^2-k+41$ is Odd number is True

We need to prove it is True for $n=k+1$

  $\begin{array}{l}{n}^2-n+41={\left(k+1\right)}^2-\left(k+1\right)+41\\ \Rightarrow n^2-n+41=k^2+2k+1-k-1+41\\ \Rightarrow n^2-n+41=\left(k^2-k+41\right)+2k\end{array}$

Here from our assumption $k^2-k+41$ is an odd number and

  $2k$ is an even number

Sum of an odd number with an even number is always an odd number.

So

  $n^2-n+41$ for $n=k+1$ is an Odd number.

And hence the proof by Mathematical induction.