Progression (Sequence and Series)HardBloom L3
Question
$\sum_{r = 1}^{99}\mspace{2mu} r!\left( r^{2} + r + 1 \right)$ is equal to
Options
A.$102! - 100$ !
B.$100(100!) - 1$
C.$99(100!) - 1$
D.$100(99!) - 1$
Solution
Sol. [Key Concept: Telescoping factorial sum: write $r \cdot \ r! = (r + 1)! - r!$ (and similar) so most terms cancel.]
$$\sum_{r = 1}^{99}\mspace{2mu} r!\left( (r + 1)^{2} - r \right) = \sum_{r = 1}^{99}\mspace{2mu}((r + 1)(r + 1)! - rr!) = 100(100!) - 1$$
Create a free account to view solution
View Solution FreeMore Progression (Sequence and Series) Questions
The A.M. of two numbers is 34 and GM is 16, the numbers are-...Let $x,y,z$ are distinct positive integers and $m = \left( \frac{x^{2} + y^{2} + z^{2}}{x + y + z} \right)^{(x + y + z)}...The mean of a set of observations is . If each observation is divided by α, α ≠ 0, and then is increased...10th term of the progression − 4 − 1 + 2 + 5 +.... is -...In the following two A.P.′s how many terms are identical ? 2,5,8,11.... to 60 terms, 3,5,7, ..... 50 terms...