Question

How can we prove using mathematical induction that ${{n}^{2}}+n+1$ is an odd number if $n$ is a natural number ?

Answer 1

Hint: Mathematical Induction is a mathematical proof technique. It basically involves three steps. First we check whether the given equation holds good for $1$. If it does, then we assume that whatever condition is given in the question also holds good for some $k$ which also belongs to natural numbers. Now, we have to prove that condition is also true for $k+1$ to the given equation in the question. Let us use these and prove that ${{n}^{2}}+n+1$ is odd.

Complete step by step solution:
Let us say that $P\left( n \right)={{n}^{2}}+n+1$ where $n$ is a natural number .
Now let us try to see what the value of $P\left( 1 \right)$ is . To get the value of $P\left( 1 \right)$, all we have to do is plug-in the value of $n$ as $n=1$
$\begin{align}
  & \Rightarrow P\left( n \right)={{n}^{2}}+n+1 \\
 & \Rightarrow P\left( 1 \right)=1+1+1=3 \\
\end{align}$
$3$ is an odd number. $P\left( 1 \right)$holds true.
Now we have to assume that ${{n}^{2}}+n+1$ is an odd number if $n$ is a natural number holds true for some $k$ where $k$ belongs to natural numbers.
Now, let us write the equation for that. To get the value of $P\left( k \right)$, all we have to do is plug-in the value of $n$ as $n=k$.
\[\begin{align}
  & \Rightarrow P\left( n \right)={{n}^{2}}+n+1 \\
 & \Rightarrow P\left( k \right)={{k}^{2}}+k+1 \\
\end{align}\]
\[{{k}^{2}}+k+1\]is an odd number .
Now let us prove \[P\left( k+1 \right)\] is also true.
To get the value of $P\left( k+1 \right)$, all we have to do is plug-in the value of $n$ as $n=k+1$.
\[\begin{align}
  & \Rightarrow P\left( n \right)={{n}^{2}}+n+1 \\
 & \Rightarrow P\left( k+1 \right)={{\left( k+1 \right)}^{2}}+k+1+1 \\
 & \Rightarrow P\left( k+1 \right)={{k}^{2}}+2k+1+k+1+1 \\
\end{align}\]
Now let us group ${{k}^{2}},k,1$ together.
Upon grouping, we get the following :
\[\begin{align}
  & \Rightarrow P\left( n \right)={{n}^{2}}+n+1 \\
 & \Rightarrow P\left( k+1 \right)={{\left( k+1 \right)}^{2}}+k+1+1 \\
 & \Rightarrow P\left( k+1 \right)={{k}^{2}}+2k+1+k+1+1 \\
 & \Rightarrow P\left( k+1 \right)={{k}^{2}}+k+1+2k+2 \\
 & \Rightarrow P\left( k+1 \right)={{k}^{2}}+k+1+2\left( k+1 \right) \\
\end{align}\]
\[{{k}^{2}}+k+1\]is nothing $P\left( k \right)$. And we assumed that $P\left( k \right)$ is an odd number. And the other term $2\left( k+1 \right)$ is clearly an even number.
We all know that the sum of an odd number and an even number is always odd.
So $P\left( k+1 \right)$ is odd.
Thus, \[P\left( k+1 \right)\] is true, whenever $P\left( k \right)$ is true.
$\therefore $ Hence, proved.

Note: These kinds of proofs may take little more time in the exam. These kinds of sums are to be practiced a lot so that you think of the easiest way possible to complete in the exam. It is very important to remember the three basic rules of mathematical induction. We can also solve sums using different methods. We can solve using proof by contradiction. In this, we first assume whatever condition is given to us to be false. And then proceed to do the sum. We end at the same result.