Question

The sum of n terms of 1^ 2 ,3^ 2 ,5^ 2 ,7^ 2 ,...., is (b) (n / 2)(4n ^ 2 - 1) a)(n/3)(4n^ 2 -1)​

Answer 1

Step-by-step explanation:

We're asked to find the simplified form of,

$\longrightarrow S=\underbrace{1^2+3^2+5^2+7^2+\,\dots}_{\textrm{ n terms}}$

The n'th term of the sequence $1^2,\ 3^2,\ 5^2,\ 7^2,\,\dots$ is $(2n-1)^2.$ So,

$\longrightarrow S=1^2+3^2+5^2+\,\dots\,+(2n-1)^2$

We know the following sum,

$\longrightarrow 1^2+2^2+3^2+\,\dots\,+n^2=\dfrac{n(n+1)(2n+1)}{6}$

This is expression for the sum of squares of first 'n' natural numbers. In case of '2n' natural numbers,

$\longrightarrow 1^2+2^2+3^2+\,\dots\,+(2n)^2=\dfrac{2n(2n+1)(4n+1)}{6}$

We separate odd squares and even squares in the LHS.

$\begin{aligned}\longrightarrow\ \ &\Big(1^2+3^2+5^2+\,\dots\,+(2n-1)^2\Big)&\\+\ \ &\Big(2^2+4^2+6^2+\,\dots\,+(2n)^2\Big)&=\dfrac{2n(2n+1)(4n+1)}{6}\end{aligned}$

We can take 4 common from the sum of even squares.

$\begin{aligned}\longrightarrow\ \ &\Big(1^2+3^2+5^2+\,\dots\,+(2n-1)^2\Big)&\\+\ \ &4\Big(1^2+2^2+3^2+\,\dots\,+n^2\Big)&=\dfrac{2n(2n+1)(4n+1)}{6}\end{aligned}$

Now the sums in LHS are, as written above,

$1^2+3^2+5^2+\,\dots\,+(2n-1)^2=S$

$1^2+2^2+3^2+\,\dots\,+n^2=\dfrac{n(n+1)(2n+1)}{6}$

So,

$\longrightarrow S+\dfrac{4n(n+1)(2n+1)}{6}=\dfrac{2n(2n+1)(4n+1)}{6}$

$\longrightarrow S=\dfrac{2n(2n+1)(4n+1)}{6}-\dfrac{4n(n+1)(2n+1)}{6}$

$\longrightarrow S=\dfrac{2n+1}{6}\Big[2n(4n+1)-4n(n+1)\Big]$

$\longrightarrow S=\dfrac{2n+1}{6}\Big[4n^2-2n\Big]$

$\longrightarrow S=\dfrac{2n(2n-1)(2n+1)}{6}$

$\longrightarrow\underline{\underline{S=\dfrac{n}{3}\,(4n^2-1)}}$

This is the expression for the sum of squares of first 'n' odd natural numbers.