Question

By method of mathematical induction, the value of $1^2+3^2+5^2\cdots +(2n-1{)}^2=\frac{n}{x}(2n-1\left)\right(2n+1)$ is true for all $n\in N$
Then $x$ is:

• A. $2$
• B. $4$
• C. $6$
• D. $3$

Answer 1

The correct option is D $3$

Let $P\left(n\right)=1^2+3^2+5^2\cdots +(2n-1{)}^2\forall n\in N$
So, it is also true for $n=1,$
$P\left(1\right)=1$
From $RHS,$ we have $P\left(1\right)=\frac{1}{x}1.3=1$
Hence on comparing we have $x=3$
Let, $P\left(k\right):1^2+3^2+5^2\cdots +(2n-1{)}^2\forall n\in N$ is true for $k=n$
now check at $k=n+1,$
$P(n+1):1^2+2^2+3^2+\cdots +(2k-1{)}^2+(2k+1{)}^2=\frac{n}{3}[4n^2-1]+(2n+1{)}^2=\frac{1}{3}[4n^3-n+12n^2+12n+3]=\frac{(n+1)}{3}[4n^2+8n+3]=\frac{(n+1)}{3}[4(n+1{)}^2-1]$
So, $P(n+1)$ is also true.