Question

find the sum of the first 'n' odd natural numbers. Hence find 1 + 3 + 5 + ... + 101.

Answer 1

The first n odd numbers are 1, 3, 5, 7,…
Now, we observe that the numbers are in an A.P.
Here,
First term, a = 1
Common difference, d = t₂ – t₁ = 3 – 1 = 2
We know that the sum of n terms of an A.P. is $Sn=\frac{n}{2}\left[2a+(n-1)d\right]$.

Thus, we get the sum of the first n odd numbers as:
$$\begin{gathered} S_n=\frac{n}{2}\left[2\left(1\right)+(n-1)\left(2\right)\right] \\ =\frac{n}{2}\left[2+2n-2\right] \\ =\frac{n}{2}\times 2n \\ =n^2 \end{gathered}$$

Now,
To find the sum of 1 + 3 + 5 + … + 101, we need to find the number of the terms.
Let the number of the terms be n.
We have:
t_n = 101
We know that the n^th term of an A.P. is t_n = a + (n – 1)d.
Thus, we get:
101 = 1 + (n – 1)2
$\Rightarrow$101 – 1 = 2(n – 1)
$\Rightarrow$100 = 2(n – 1)
$\Rightarrow$(n – 1) = 50
$\Rightarrow$n = 50 + 1 = 51
We have the sum of the first n odd numbers as $Sn=n^2$.
Thus, we have:
$S_{51}=(51{)}^2=2601$