CAT 2018

Slot 2

QA

Question & Solution

S = 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99

Nth term of the series can be written as $T_{n} = (4n+3)*(4n+7)$

Last term, (4n+3) = 95 i.e. n = 23

$\sum_{n=1}^{n=23} (4n+3)*(4n+7)$

$\Rightarrow$ $\sum_{n=1}^{n=23}16n^2+40n+21$

$\Rightarrow$ $16*\dfrac{23*24*47}{6}+40*\dfrac{23*24}{2}+21*23$

$\Rightarrow$ $80707$