Question

write the sum of following without actually calculating 1 + 3 + 5 +....+19​

Answer 1

Hope it's Helpful for you....... .

Answer 2

This series contains positive odd integers. First we have to compare them with respect to consecutive natural numbers.

Consider an infinite series of odd integers.

1 + 3 + 5 + ......∞

We can note each term as the following.

1 = 2 - 1 = (2 • 1 - 1)

3 = 4 - 1 = (2 • 2 - 1)

5 = 6 - 1 = (2 • 3 - 1)

So we can express the nth term of the series as 2n - 1.

Now, consider the series up to n terms.

1 + 3 + 5 + ...... + (2n - 1)

Simplifying this using Sigma notation...

$\begin{aligned}&1+3+5+......+(2n-1)\\ \\ \Longrightarrow\ \ &\sum_{k=1}^{n}2k-1\\ \\ \Longrightarrow\ \ &2\sum_{k=1}^{n}k-\sum_{k=1}^n1\\ \\ \Longrightarrow\ \ &2\left(\frac{n(n+1)}{2}\right)-(\underbrace{1+1+1+...+1}_{n})\\ \\ \Longrightarrow\ \ &n(n+1)-n\\ \\ \Longrightarrow\ \ &n(n+1-1)\\ \\ \Longrightarrow\ \ &n^2\end{aligned}$

So, we got a general formula.

1 + 3 + 5 + ... + (2n - 1) = n²

Now, come to the question.

1 + 3 + 5 +....+ 19

Here,

2n - 1 = 19

2n = 20

n = 10

So,

1 + 3 + 5 +....+ 19 = 10² = 100