What is the value of 1*1! + 2*2! + 3*3! + ... + n*n! ?

Question

What is the value of  1*1! + 2*2! + 3*3! + ... + n*n!  ?

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

gmatophobia · Accepted Answer

A quick way to solve such questions is via use of options and via substitution -

Assume n = 3

1*1! + 2*2! + 3*3!  = 1 + 4 + 18 = 23

Answer choice elimination

(A) n! +1

3! + 1 = 7 →  Eliminate

(B) (n + 1)!

(3 + 1) = 4! = 24 →  Eliminate

(C) (n + 1)!-1

(3 + 1) - 1 = 4!-1 = 24-1 = 23 →  Keep

(D) (n + 1)!+1

(3 + 1) + 1 = 4!+1 = 24+1 = 25 →  Eliminate

(E) n! + 3

3! + 3 = 6 + 3 = 9 →  Eliminate

Option C

Oppenheimer1945 · Answer

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

The sum telescopes to (n+1)!-1

Posted from my mobile device

TheVDR · Answer

n = 1 should give 1 as the answer. Only option C does that.   .

Else, the form is n*n!,

Substitute n = n + 1 - 1,

=>   * n!
=> (n+1)*n! - n!
=> (n+1)! - n!

Making a series out of it,
(2! - 1!) + (3! - 2!) + (4! - 3!) + ..... +   +

=> We are left with -1! + (n+1)!, while all other terms cancel each other.

Therefore, the answer is C. (n+1)! - 1

btsaami · Answer

Best way is to consider the series for smaller no. n=1, 2 or 3.

Method 2:

1*1! + 2*2! + ---+n*n!
This can be arranged and written as
(2*1! - 1!) + (3*2!- 2!) + (4*3!- 3!) + --- +((n+1)*n!- n!)

2!- 1! + 3! - 2! +4! - 3! + ------+ (n+1)!- n!
We see that values are getting cancelled other than 1 and (n+1)!

Therefore, we get (n+1)!- 1

bumpbot · Answer

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What is the value of 11! + 22! + 33! + ... + nn! ?

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