Question

Show by induction, that \forall n\ge 1,$$$${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)?

Answer 1

This problem is based on an induction method, which is used to prove a statement where n represents the set of all natural numbers. It involves two steps, first one is the statement which is true for n = 1 and n = 2 and the second is true for n = k and prove that if it is true for n = k, then it is also true for n = k+1. We can do it step by step to prove the given expression.

Complete step by step solution:
We have to show that,
${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)$
We can now write the first step,
For ${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)$, we can take n = 1, we get
$\Rightarrow {{1}^{2}}=\dfrac{1}{3}\left( 4\times {{1}^{2}}-1 \right)=1$
For ${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)$, we can take n = 2, we get
$\Rightarrow {{1}^{2}}+{{3}^{2}}=\dfrac{2}{3}\left( 4\times {{2}^{2}}-1 \right)=\dfrac{2}{3}\times 15=10$
Hence, we can see that the given statement is true for n = 1 and n = 2.
We can now write the second step,
For ${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)$, we can take n = k, we get
$\Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2k-1 \right)}^{2}}=\dfrac{k}{3}\left( 4{{k}^{2}}-1 \right)$
For ${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)$, we can take n = k+1, we get

& \Rightarrow {{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2k-1 \right)}^{2}}+{{\left( 2k+1 \right)}^{2}} \\\ & =\dfrac{k}{3}\left( 4{{k}^{2}}-1 \right)+4{{k}^{2}}+4k+1 \\\ & =\dfrac{k}{3}\left( 4{{k}^{2}}+8k+4-1 \right)-\dfrac{k}{3}\left( 8k+4 \right)+4{{k}^{2}}+4k+1 \\\ & =\dfrac{k}{3}\left( 4{{k}^{2}}+8k+4-1 \right)-\dfrac{8k}{3}-\dfrac{4k}{3}+4{{k}^{2}}+4k+1 \\\ & =\dfrac{k}{3}\left( 4{{k}^{2}}+8k+4-1 \right)-\dfrac{1}{3}\left( 4{{k}^{2}}+8k+4-1 \right)-\dfrac{8{{k}^{2}}}{3}-\dfrac{4k}{33}+4{{k}^{2}}+4k+1 \\\ & =\dfrac{k+1}{3}\left( 4{{\left( k+1 \right)}^{2}}-1 \right) \\\ \end{aligned}$$ Hence, we can see that the given statement is true for n = k+1 if it is true for n = k. Therefore, $${{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{\left( 2n-1 \right)}^{2}}=\dfrac{n}{3}\left( 4{{n}^{2}}-1 \right)$$ is true for all values of $$n\in \mathbb{N}$$. **Note:** We should know the concept of the induction method to prove these types of problems. We should concentrate on the part, where we substitute n = k+1 as it is a bit complicated. We should also calculate the steps where we will make mistakes in the sign part.