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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

Here, the expression on the left side of the given equation is an arithmetic sequence. We will first find the first term, last terms, and the common difference. We will then substitute these values in the formula of the sum of nn terms of an AP and simplify it further to get the required answer.

Formula Used:
Sum of nn terms of an AP =n2(a+an) = \dfrac{n}{2}\left( {a + {a_n}} \right), where aa is the first term, an{a_n} is the last term and nn represents the total number of terms in the AP.

Complete step by step solution:
We can observe that the given sequence is an arithmetic progression. Here,
First term, a=1a = 1
Common difference, d=3−1=5−3=2d = 3 - 1 = 5 - 3 = 2
The last term, an=(2n−1){a_n} = \left( {2n - 1} \right)
Sum of nn terms of an AP =n2(a+an) = \dfrac{n}{2}\left( {a + {a_n}} \right)
Hence, we can write the LHS as:
1+3+5+.....+(2n−1)=n2(1+2n−1)=n2(2n)1 + 3 + 5 + ..... + \left( {2n - 1} \right) = \dfrac{n}{2}\left( {1 + 2n - 1} \right) = \dfrac{n}{2}\left( {2n} \right)
Here, substituting n=1n = 1 in the LHS and RHS respectively, we get,
LHS =1+3+5+.....+(2n−1)=n2(2n)=12(2×1)=1 = 1 + 3 + 5 + ..... + \left( {2n - 1} \right) = \dfrac{n}{2}\left( {2n} \right) = \dfrac{1}{2}\left( {2 \times 1} \right) = 1
RHS =n2=12=1 = {n^2} = {1^2} = 1
Hence, LHS == RHS =1 = 1
Now, substituting n=kn = k such that,
1+3+5+.....+(2k−1)=k21 + 3 + 5 + ..... + \left( {2k - 1} \right) = {k^2}
We need to prove that the result is also true for n=k+1n = k + 1,
Hence, adding (2k+1)\left( {2k + 1} \right) on both sides, we get,
1+3+5+.....+(2k−1)+(2k+1)=k2+2k+11 + 3 + 5 + ..... + \left( {2k - 1} \right) + \left( {2k + 1} \right) = {k^2} + 2k + 1
Using the identity (a+b)3=a2+2ab+b2{\left( {a + b} \right)^3} = {a^2} + 2ab + {b^2} in the RHS,
⇒1+3+5+.....+(2k−1)+(2k+1)=(k+1)2\Rightarrow 1 + 3 + 5 + ..... + \left( {2k - 1} \right) + \left( {2k + 1} \right) = {\left( {k + 1} \right)^2}
Hence, the result is true for n=k+1n = k + 1
Therefore, by principle of mathematical induction or POMI, the result is true for all n∈Zn \in Z
Therefore,
1+3+5+.....+(2n−1)=n21 + 3 + 5 + ..... + \left( {2n - 1} \right) = {n^2}
Hence, proved

Note:
An Arithmetic Progression or A.P. is a sequence in which the difference between two consecutive terms is the same. This difference is known as the common difference and we find it by subtracting each term by its preceding term respectively. Arithmetic progressions are also used in real life such as adding the same amount as our pocket money in our money bank. Since, we add the same amount each time, that amount or that pocket money will become our common difference in this case.