Question

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

Answer 1

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

Complete step by step solution:
We will first check if the result is true for \[n = 1\].
On the L.H.S., we will get 1.
On the R.H.S. we will get \[{\left( 1 \right)^2} = 1\].
As \[{\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = {\rm{R}}{\rm{.H}}{\rm{.S}}{\rm{.}}\]
 The result is true for \[n = 1\].
We will assume, that the result is true for \[n = k\]:
\[\Rightarrow 1 + 3 + 5 + ... + \left( {2k - 1} \right) = {k^2}\]
We will prove the result is true for \[n = k + 1\].
\[\Rightarrow {\rm{L}}{\rm{.H}}{\rm{.S}}{\rm{.}} = 1 + 3 + 5 + ... + \left( {2k - 1} \right) + \left( {2\left( {k + 1} \right) - 1} \right)\]
From step 2, we know that \[1 + 3 + 5 + ... + \left( {2k - 1} \right) = {k^2}\]. We will substitute the value of \[1 + 3 + 5 + ... + \left( {2k - 1} \right)\] in the expression on L.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 \[n = k + 1\], The result is true for all \[n \in \mathbb{N}\] (Natural Numbers).

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