Question

10. Sum of the first n odd natural numbers is​

Answer 1

TO FIND :–

• Sum of n odd number = ?

SOLUTION :–

• Series of odd numbers –

$\\ \sf \implies 1,3,5,7,9,.........n \: terms$

• We know that sum of n terms –

$\\ \longrightarrow \: \: \large{ \boxed{ \sf S_{n} = \dfrac{n}{2}[2a + (n - 1)d]}}$

• Here –

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \sf a = first \: \: term =1$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: \sf d = common \: \: difference =2$

• So that –

$\\ \implies \sf S_{n} = \dfrac{n}{2}[2(1) + (n - 1)2]$

$\\ \implies \sf S_{n} = \dfrac{n}{2}[2 +2n - 2]$

$\\ \implies \sf S_{n} = \dfrac{n}{2}(2n)$

$\\ \implies \sf S_{n} = \dfrac{ \cancel2 {n}^{2} }{ \cancel2}$

$\\ \implies \large{ \boxed{ \sf S_{n} = {n}^{2}}}$

• Hence , Sum of n odd terms is n² .

Answer 2

Answer:

The sum of first n odd natural numbers is always n²