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Question: Prove that \[1 + 3 + 5 + ... + \left( {2n - 1} \right) = {n^2}\]....

Prove that 1+3+5+...+(2n−1)=n21 + 3 + 5 + ... + \left( {2n - 1} \right) = {n^2}.

Explanation

Solution

We will prove the result using the Principle of Mathematical Induction. We will check if the result is true for n=kn = k. Then, we will assume that the result is true for n=kn = k. We will prove that the result is true for n=k+1n = k + 1. We will use the formula for the sum of first 2n2n natural numbers. We will subtract the sum of first nn even numbers from it. We will obtain the sum of the first nn odd numbers.

Complete step by step solution:
We will first check if the result is true for n=1n = 1.
On the L.H.S., we will get 1.
On the R.H.S. we will get (1)2=1{\left( 1 \right)^2} = 1.
As L.H.S.=R.H.S.{\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}}
The result is true for n=1n = 1.
We will assume, that the result is true for n=kn = k:
⇒1+3+5+...+(2k−1)=k2\Rightarrow 1 + 3 + 5 + ... + \left( {2k - 1} \right) = {k^2}
We will prove the result is true for n=k+1n = k + 1.
⇒L.H.S.=1+3+5+...+(2k−1)+(2(k+1)−1)\Rightarrow {\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = 1 + 3 + 5 + ... + \left( {2k - 1} \right) + \left( {2\left( {k + 1} \right) - 1} \right)
From step 2, we know that 1+3+5+...+(2k−1)=k21 + 3 + 5 + ... + \left( {2k - 1} \right) = {k^2}. We will substitute the value of 1+3+5+...+(2k−1)1 + 3 + 5 + ... + \left( {2k - 1} \right) in the expression on L.H.S.
1+3+5+...+(2k−1)+(2(k+1)−1)=k2+2k+2−1 ⇒k2+2k+1 ⇒(k+1)2 ⇒R.H.S.\begin{array}{l}1 + 3 + 5 + ... + \left( {2k - 1} \right) + \left( {2\left( {k + 1} \right) - 1} \right) = {k^2} + 2k + 2 - 1\\\ \Rightarrow {k^2} + 2k + 1\\\ \Rightarrow {\left( {k + 1} \right)^2}\\\ \Rightarrow {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}}\end{array}

As the result is true for n=k+1n = k + 1, The result is true for all n∈Nn \in \mathbb{N} (Natural Numbers).

Note:
We can also prove the result with an alternate method. We know that the sum of the first nn natural numbers is n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2} .
1+2+3+...+n=n(n+1)21 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2} .
The sum of first 2n2n natural numbers will be:
1+2+3+4+5+...+(2n−1)+2n=(2n)(2n+1)2 ⇒1+2+3+4+5+...+(2n−1)+2n=n(2n+1) ⇒1+2+3+4+5+...+(2n−1)+2n=2n2+n\begin{array}{l}1 + 2 + 3 + 4 + 5 + ... + \left( {2n - 1} \right) + 2n = \dfrac{{\left( {2n} \right)\left( {2n + 1} \right)}}{2}\\\ \Rightarrow 1 + 2 + 3 + 4 + 5 + ... + \left( {2n - 1} \right) + 2n = n\left( {2n + 1} \right)\\\ \Rightarrow 1 + 2 + 3 + 4 + 5 + ... + \left( {2n - 1} \right) + 2n = 2{n^2} + n\end{array}
We will separate the odd numbers and even numbers and write the sum of the first 2n2n numbers as sum of even numbers and odd numbers:
2n2+n=1+2+3+4+...+(2n−1)+2n ⇒2n2+n=1+3+5+...+(2n−1)⏟+2+4+6+...+2n⏟\begin{array}{l}2{n^2} + n = 1 + 2 + 3 + 4 + ... + \left( {2n - 1} \right) + 2n\\\ \Rightarrow 2{n^2} + n = \underbrace {1 + 3 + 5 + ... + \left( {2n - 1} \right)}_{} + \underbrace {2 + 4 + 6 + ... + 2n}_{}\end{array}
We will take 2 common from the set of even numbers:
2n2+n=1+3+5+...+(2n−1)⏟+2(1+2+3+...+n)⏟2{n^2} + n = \underbrace {1 + 3 + 5 + ... + \left( {2n - 1} \right)}_{} + \underbrace {2\left( {1 + 2 + 3 + ... + n} \right)}_{}
We will substitute n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2} for 1+2+3+...+n1 + 2 + 3 + ... + n in the above equation:
2n2+n=1+3+5+...+(2n−1)⏟+2×n(n+1)2⏟ ⇒2n2+n=1+3+5+...+(2n−1)+n(n+1) ⇒2n2+n=1+3+5+...+(2n−1)+n2+n\begin{array}{l}2{n^2} + n = \underbrace {1 + 3 + 5 + ... + \left( {2n - 1} \right)}_{} + \underbrace {2 \times \dfrac{{n\left( {n + 1} \right)}}{2}}_{}\\\ \Rightarrow 2{n^2} + n = 1 + 3 + 5 + ... + \left( {2n - 1} \right) + n\left( {n + 1} \right)\\\ \Rightarrow 2{n^2} + n = 1 + 3 + 5 + ... + \left( {2n - 1} \right) + {n^2} + n\end{array}
We will subtract n2+n{n^2} + n from both sides:
2n2+n−n2+n=1+3+5+...+(2n−1) ⇒n2=1+3+5+...+(2n−1)\begin{array}{l}2{n^2} + n - {n^2} + n = 1 + 3 + 5 + ... + \left( {2n - 1} \right)\\\ \Rightarrow {n^2} = 1 + 3 + 5 + ... + \left( {2n - 1} \right)\end{array}
We can see that the result of the first nn odd numbers is equal to n2{n^2}.
Hence, we have proved the desired result.