(C) Let the sum be $S_n = 1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$.We can rewrite the $k$-th term as $k(k!) = ((k + 1) - 1)k! = (k + 1)! - k!$.Now,we su

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(C) Let the sum be $S_n = 1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$.We can rewrite the $k$-th term as $k(k!) = ((k + 1) - 1)k! = (k + 1)! - k!$.Now,we sum these terms from $k = 1$ to $n$:$S_n = \sum_{k=1}^{n} ((k + 1)! - k!)$$S_n = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n + 1)! - n!)$This is a telescoping series where intermediate terms cancel out.$S_n = (n + 1)! - 1!$$S_n = (n + 1)! - 1$.

Competitive Quantitative Aptitude Progression and Sequence Progression and Sequence

Difficult

The sum $1(1!) + 2(2!) + 3(3!) + ....+n (n!)$ equals

A
$3(n!) + n - 3$
B
$(n + 1)! - (n - 1)!$
C
$(n + 1)! - 1$
D
$2(n!) - 2n - 1$

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Progression and Sequence

If $f(x)$ is a function satisfying $f(x + y) = f(x)f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x = 1}^n f(x) = 120$,then the value of $n$ is

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The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

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Suppose $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ If $a < b < c$ and $a + b + c = \frac{3}{2}$,then the value of $a$ is

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The solution of the equation $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$ is

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The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ will be

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The sum $1(1!) + 2(2!) + 3(3!) + ....+n (n!)$ equals

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If $f(x)$ is a function satisfying $f(x + y) = f(x)f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x = 1}^n f(x) = 120$,then the value of $n$ is

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