Question

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

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

Answer 1

The correct option is D 2220

${45}^2-{43}^2+{44}^2-{42}^2+{43}^2-{41}^2+{42}^2-{40}^2+\cdots \text{15 terms}$

$=(45+43)(45-43)+(44+42)(44-42)+(43+41)(43-41)+....\text{15 terms}$

$=(45+43)\left(2\right)+(44+42)\left(2\right)+(43+41)\left(2\right)+....\text{15 terms}$

$=2\left[\right(45+44+43+...\text{15 terms})+(43+42+41+...\text{15 terms}]$

$=2\times \frac{15}{2}[2\times 45+(15-1\left)\right(-1\left)\right]+2\times \frac{15}{2}[2\times 43+(15-1\left)\right(-1\left)\right]$

$=15(90-14)+15(86-14)$

$=15\left(76\right)+15\left(72\right)$

$=2220$