A number theory problem by Vibha Sharma

Number Theory Level 2

1 ⋅ 1 ! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + … + n ⋅ n ! = ? \large 1\cdot1 ! + 2\cdot2! + 3\cdot3!+ \ldots + n\cdot n! = \ ?

( n + 1 ) ! − 1 (n+1) ! -1 ( n + 1 ) ! + 1 ( n+1 )! + 1 ( n + 1 ) ! (n+1) ! None of these choices

Vibha Sharma by Vibha Sharma , 26, India

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3 solutions

Raj Rajput
Aug 25, 2015

8 Helpful 0 Interesting 0 Brilliant 0 Confused
Aadeesh Shastry
Aug 24, 2015

Let us use the fact that (n+1) \times n! equals (n+1)! Each term in the summation can be written as n \times n! for some natural number. If we were to subtract n \times n! From (n+1) \ times n! We would get n! after applying the distributive property. This means that for every term i \times i! In the summation, we can replace it with (i+1)! - i!. This summation telescopes yielding the answer (n+1)! - 1

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Mohit Gupta
Aug 25, 2015

A pretty easy way to solve the question assume n=1,2,3,4 and solve the question by trying the options :p

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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