Question

Find the sum of 100 terms of the series $1\left(3\right)+3\left(5\right)+5\left(7\right)+\cdots \cdots$

• A. 1353300
• B. 1353400
• C. 1353200
• D. 1353100

Answer 1

The correct option is A 1353300

Given

$S_n=1\left(3\right)+3\left(5\right)+5\left(7\right)+......$

It can be written as:

$S_n=\sum _{n=1}^{100}(2n-1)(2n+1)$

$S_n=\sum _{n=1}^{100}(4n^2+2n-2n-1)$

$S_n=\sum _{n=1}^{100}(4n^2-1)$

$S_n=\sum _{n=1}^{100}4n^2-\sum _{n=1}^{100}1$

$S_n=\frac{4\times 100\left(101\right)\left(201\right)}{6}-100$

$S_n=1353400-100$

$S_n=1353300$

Remember

$\sum _{n=1}^n{n}^2=\frac{n(n+1)(2n+1)}{6}$

$\sum _{n=1}^{n}1=n$

$\sum _{n=1}^{n}n=\frac{n(n+1)}{2}$.