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

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=(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=(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)&=(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)&=(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=(n(n+1)(2n+1))/(6)`

So,

`\longrightarrow S+(4n(n+1)(2n+1))/(6)=(2n(2n+1)(4n+1))/(6)`

`\longrightarrow S=(2n(2n+1)(4n+1))/(6)-(4n(n+1)(2n+1))/(6)`

`\longrightarrow S=(2n+1)/(6)\Big[2n(4n+1)-4n(n+1)\Big]`

`\longrightarrow S=(2n+1)/(6)\Big[4n^2-2n\Big]`

`\longrightarrow S=(2n(2n-1)(2n+1))/(6)`

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

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