Question

The sum of the series
${45}^2-{43}^2+{44}^2-{42}^2+{43}^2-{41}^2+{42}^2-{40}^2+\cdots 15terms$ is

• A. 1900
• B. 1290
• C. 1650
• D. 2220

Answer 1

The correct option is D

2220

Given, series
${45}^2-{43}^2+{44}^2-{42}^2+{43}^2-{41}^2+{42}^2-{40}^2+\cdots 15terms$
$\Rightarrow ({45}^2-{43}^2)+({44}^2-{42}^2)+({43}^2-{41}^2)+\cdots 15terms$
$=(45+43)(45-43)+(44+42)(44-42)+(43+41)(43-41)+....15terms$
$=(45+43)\left(2\right)+(44+42)\left(2\right)+(43+41)\left(2\right)+....15terms$
$=2\left[\right(45+44+43+...15terms)+(43+42+41+...15terms\left)\right]$
$=2\times \frac{15}{2}[2\times 45+(15-1\left)\right(-1\left)\right]+2\times \frac{15}{2}[2\times 43+(15-1\left)\right(-1\left)\right]$
$=15(90-14)+15(86-14)$
$=15\left(76\right)+15\left(72\right)$
$=2220$