Question

ILLUSTRATIVE EXAMPLES
Find the sum to n terms of the series 1^2 + 3^2 +5^2 + ... to n terms.
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Answer 1

Given series is,

$1^2+3^2+5^2+...$

This is sum of squares of first consecutive positive odd integers.

The series is taken upto  n  terms, so that the  nth  term will be 2n - 1.

Hence the series will be,

$1^2+3^2+5^2+...+(2n-1)^2$

This sum can be written in sigma notation as,

$\displaystyle 1^2+3^2+5^2+...+(2n-1)^2=\sum_{k=1}^{n}(2k-1)^2$

Now, we can find the general formula for this series.

$\begin{aligned}&\sum_{k=1}^{n}(2k-1)^2\\ \\ \Longrightarrow\ \ &\sum_{k=1}^{n}4k^2-4k+1\\ \\ \Longrightarrow\ \ &4\sum_{k=1}^nk^2-4\sum_{k=1}^nk+\sum_{k=1}^n1\\ \\ \Longrightarrow\ \ &4\left(\frac{n(n+1)(2n+1)}{6}\right)-4\left(\frac{n(n+1)}{2}\right)+n\\ \\ \Longrightarrow\ \ &\frac{2n(n+1)(2n+1)}{3}-2n(n+1)+n\\ \\ \Longrightarrow\ \ &2n(n+1)\left(\frac{2n+1}{3}-1\right)+n\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &2n(n+1)\left(\frac{2(n-1)}{3}\right)+n\\ \\ \Longrightarrow\ \ &\large \bigg(\frac{4(n-1)(n+1)}{3}+1\bigg)n\end{aligned}$

So, we got a general formula.

Let's check whether it's true.

$\textsf{Let \ n=1. }\\ \\ 1^2=\bold{1}\\ \\ \left(\dfrac{4\cdot 0\cdot 2}{3}+1\right)1=0+1=\bold{1}\\ \\ \\ \\ \textsf{Let \ n=2. }\\ \\ 1^2+3^2=1+9=\bold{10}\\ \\ \left(\dfrac{4\cdot 1\cdot 3}{3}+1\right)2=8+2=\bold{10}\\ \\ \\ \\ \textsf{Let \ n=3. }\\ \\ 1^2+3^2+5^2=1+9+25=\bold{35}\\ \\ \left(\dfrac{4\cdot 2\cdot 4}{3}+1\right)3=32+3=\bold{35}$

$\textsf{Let \ n=k. }\\ \\ \textsf{Assuming \ 1^2+3^2+5^2+...+(2k-1)^2=\left(\dfrac{4(k-1)(k+1)}{3}+1\right)k }\\ \\ \\ \\ \textsf{Let \ n=k+1. }\\ \\ \begin{aligned}&1^2+3^2+5^2+...+(2(k+1)-1)^2\\ \\ \Longrightarrow\ \ &1^2+3^2+5^2+...+(2k-1)^2+(2k+1)^2\\ \\ \Longrightarrow\ \ &\left(\frac{4(k-1)(k+1)}{3}+1\right)k+(2k+1)^2\\ \\ \Longrightarrow\ \ &\frac{4k(k-1)(k+1)}{3}+k+4k^2+4k+1\\ \\ \Longrightarrow\ \ &\frac{4k^3-4k+3k+12k^2+12k+3}{3}\\ \\ \Longrightarrow\ \ &\frac{4k^3+12k^2+11k+3}{3}\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\frac{4k^3+12k^2+8k+3k+3}{3}\\ \\ \Longrightarrow\ \ &\frac{4k(k^2+3k+2)+3(k+1)}{3}\\ \\ \Longrightarrow\ \ &\frac{4k(k+1)(k+2)}{3}+k+1\\ \\ \Longrightarrow\ \ &\left(\frac{4k(k+2)}{3}+1\right)\bigg(k+1\bigg)\\ \\ \Longrightarrow\ \ &\left(\frac{4(k+1-1)(k+1+1)}{3}+1\right)\bigg(k+1\bigg)\\ \\ \Longrightarrow\ \ &\left(\frac{4(n-1)(n+1)}{3}+1\right)n\end{aligned}$

Yes, it's true!