The sum r = 1 10 ( r 2 + 1 ) × ( r ! ) is equal to

Question

The sum $\sum_{r = 1}^{10} (r^{2} + 1) \times (r!)$ is equal to

Smart Achievers · Accepted Answer

(2)

We have,

$T_{r} = (r^{2} + 1 + r - r) r! = (r^{2} + r) r! - (r - 1) r!$

$\Rightarrow$ $T_{r} = r (r + 1)! - (r - 1) r!$

$∴$ $T_{1} = 1 \cdot 2! - 0$

$T_{2} = 2 \cdot 3! - 1 \cdot 2!$

$T_{3} = 3 \cdot 4! - 2 \cdot 3!$

.................................................

$T_{10} = 10 \cdot 11! - 9 \cdot 10!$

$∴$ $\sum_{r = 1}^{10} (r^{2} + 1) r! = 10 \cdot 11!$

Want Us to Email you About Special Offers & Updates?