Simplify it!

Algebra Level 3

1 2 ! + 2 3 ! + 3 4 ! + . . . . . . + n ( n + 1 ) ! \frac { 1 }{ 2! } +\frac { 2 }{ 3! } +\frac { 3 }{ 4! } +......+\frac { n }{ (n+1)! }

(n-1)! 0 (n+1)! 1+1/(n+1)! 1-(n-1)! n! 1-1/(n+1)! 1-(n+1)!

Zhaochen Xie by Zhaochen Xie , 19, Spain

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1 solution

Bala Vidyadharan
Nov 18, 2015

Simple just use the concept of telescopic sum.

Let the given expression be S

S= 1 2 ! + 2 3 ! + . . . . . . . . . + n ( n + 1 ) ! 2 − 1 2 ! + 3 − 1 3 ! + . . . . + n + 1 − 1 ( n + 1 ) ! 2 2 ! − 1 2 ! + 3 3 ! − 1 3 ! + . . . . + n + 1 ( n + 1 ) ( n ) ! − 1 ( n + 1 ) ! A l l t h e m i d d l e t e r m s g e t s c a n c e l l e d e x c e p t 1 − 1 ( n + 1 ) ! \frac { 1 }{ 2! } +\frac { 2 }{ 3! } +.........+\frac { n }{ (n+1)! } \\ \\ \frac { 2-1 }{ 2! } +\frac { 3-1 }{ 3! } +....+\frac { n+1-1 }{ (n+1)!\\ } \\ \frac { 2 }{ 2! } -\frac { 1 }{ 2! } +\frac { 3 }{ 3! } -\frac { 1 }{ 3! } +....+\frac { n+1 }{ (n+1)(n)! } \quad -\quad \frac { 1 }{ (n+1)! } \\ \\ All\quad the\quad middle\quad terms\quad gets\quad cancelled\quad except\\ \\ 1-\frac { 1 }{ (n+1)! }

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