Question

Prove by method of induction, for all \[n\in 3+7+11+..........\]to $n$ terms = \[n\left( 2n+1 \right)\] state true or false.

Answer 1

Hint: By using mathematical induction prove that the given equation is true for all positive integers. First we check the given statement by putting the values of $n$. If the statement is true for all values of $n$. Then, the given statement is true.

Complete step-by-step answer:
We have the given statement as $P(n)=3+7+11+...........\text{n terms}=n(2n+1)$
We check the above statement by putting the different values of $n$ to find whether it is true or false.
First we put $n=1$ in above statement, then
The L.H.S. of the statement will be $P(1)=3$
The R.H.S. of the statement will be $1(2\times 1+1)=3$
Hence we get that the L.H.S. = R.H.S.
It means $P(1)$ is true.
Now let us assume that the result is true for $n=k$
The series becomes $3+7+11+.........\text{k terms}=k(2k+1)$
Now let us prove this result for $n=k+1$
So, we will get the series as \[3+7+11+.............\text{(k+1) terms}=(k+1)(2k+3)\]
So, we get that the $\mathop{(k+1)}^{th}term=\left( 4k+3 \right)$
Now, L.H.S. $3+7+11+......\text{to k terms}+\left( 4k+3 \right)$
$\begin{align}
  & =k\left( 2k+1 \right)+\left( 4k+3 \right) \\
 & =2\mathop{k}^{2}+k+4k+3 \\
 & =k(2k+3)+1(2k+3) \\
 & =(k+1)(2k+3) \\
\end{align}$
= R.H.S.
Thus the result is also true for $n=k+1$
Hence, by principle of mathematical induction the given statement is true for all natural numbers.
So we have got that the answer is true.

Note: Mathematical induction is a method used to establish that a given statement is true of all natural numbers. Check the given problem by putting the different values of $n$. If the statement is satisfied the principle of mathematical induction for all values of $n$, then the statement is true. If the statement is not satisfied for any value of $n$, then the statement is false.