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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 7 of 120

Marks: +1, -0

If

a_{n} = n(n!)

a_{1} + a_{2} + a_{3} + \cdots + a_{10}

equal to?

$10! - 1$
$11! + 1$
$10! + 1$
$11! - 1$

Solution:

Concept:

Rewrite

a_n = n(n!)

as a telescoping difference:

(n+1)! - n!

.

Explanation:

Given

a_n = n(n!)

.

We express it as

a_n = (n+1)n! - n! = (n+1)! - n!

.

Substitute

n=1

to

10

:

a_1 = 2! - 1!

,

a_2 = 3! - 2!

, …,

a_{10} = 11! - 10!

.

Add all terms:

(2!-1!) + (3!-2!) + \cdots + (11!-10!)

.

All intermediate terms cancel, leaving

11! - 1! = 11! - 1

.

Answer:

11! - 1

(Option D).

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