Question

Find the sum of first n odd numbers.

Answer 1

Hint: In order to find out the sum of n odd natural numbers first we will write an sequence which will be in the form of A.P. then we will simply apply the formula of sum of n term series as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

Complete step-by-step answer:

The sequence of first n odd numbers is $1,3,5,7.............(2n-1)$
This forms an A.P. with $a = 1$ and $d = 2$
The sum of first n terms is given by
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \\\ \Rightarrow S = \dfrac{n}{2}\left( {2 \times 1 + \left( {n - 1} \right)2} \right) \\\ \Rightarrow S = \dfrac{n}{2}\left( {2 + 2n - 2} \right) \\\ \Rightarrow S = \dfrac{n}{2}\left( {2n} \right) \\\ \Rightarrow S = {n^2} \\\$
Hence, the sum of first n odd terms is ${n^2}$.

Note: In order to solve these types of problems, first of all remember the formula of arithmetic progression. Remember how to find the nth term of and A.P. Also remember how to find the sum of n terms of an A.P. Similarly learn about geometric progression and harmonic progression. This will help a lot to solve problems related to A.P, G.P and H.P.