The value of the expression 3(1!) - 4(2!) + 5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the general term of the series be $T_n = (-1)^{n-1} (n+2)n!$. We can rewrite the terms as: $T_n = (-1)^{n-1} ((n+1) + 1)n! = (-1)^{n-1} ((n+1)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let the general term of the series be $T_n = (-1)^{n-1} (n+2)n!$. We can rewrite the terms as: $T_n = (-1)^{n-1} ((n+1) + 1)n! = (-1)^{n-1} ((n+1)! + n!)$. Expanding the series: $S = (2! + 1!) - (3! + 2!) + (4! + 3!) - (5! + 4!) + \dots - (2007! + 2006!) + 2007!$. Observing the telescoping nature of the sum: $S = 2! + 1! - 3! - 2! + 4! + 3! - 5! - 4! + \dots - 2007! - 2006! + 2007!$. All terms cancel out except for $1!$ and the last term. $S = 1! = 1$.

The value of the expression $3(1!) - 4(2!) + 5(3!) - 6(4!) + \dots - 2008(2006)! + (2007)!$ is:

Similar Questions

The solution of the equation $1 + a + a^2 + a^3 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$ is given by $x$ is equal to

If $a$ is the first term of a $G.P.$,$l$ is the $n^{th}$ term,and $P$ is the product of the first $n$ terms,then $P=$

If $x, y, z$ are three positive numbers in $G.P.$,then $\frac{1 + \ln x}{2}, \frac{1 + \ln y}{4}, \frac{1 + \ln z}{8}$ are in

The number $111...1$ ($91$ times) is a:

Three numbers are in $A.P.$ whose sum is $33$ and product is $792$. The smallest number among these is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module