JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q69

MathematicsSingle correct+4 / -1Question 69 of 90

The value of $\displaystyle\sum_{r=1}^{15}r^2\left(\frac{{}^{15}C_r}{{}^{15}C_{r-1}}\right)$ is equal to :

560

680

1240

1085

Answer and worked solution

Official answer: B

$\frac{^{15}C_r}{^{15}C_{r-1}}=\frac{16-r}r$ , so the sum is $\sum_{r=1}^{15}r(16-r)=16\times120-1240=680$ .

JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 69