Evaluate-3-1-2-3-4-2-3-4-2001-1999-2000-2001-

Question Number 97136 by I want to learn more last updated on 06/Jun/20

Evaluate

\frac{3}{1! + 2! + 3!} + \frac{4}{2! + 3! + 4!} + \dots + \frac{2001}{1999! + 2000! + 2001!}

Commented by 06122004 last updated on 06/Jun/20

* Y e c h i m *

n! + (n + 1)! + (n + 2)! = n! (1 + (n + 1) + (n + 1) (n + 2)) =

= n! * {(n + 2)}^{2} ⊛

\frac{3}{1! * 3^{2}} + \frac{4}{2! * 4^{2}} + \dots + \frac{2001}{1999! * 2001^{2}} =

= \frac{1}{1! * 3} + \frac{1}{2! * 4} + \dots + \frac{1}{1999! * 2001} =

= \frac{2}{3!} + \frac{3}{4!} + \dots + \frac{2000}{2001!} =

= \frac{3 - 1}{3!} + \frac{4 - 1}{4!} + \dots + \frac{2001 - 1}{2001!} =

= \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \dots + \frac{1}{2000!} - \frac{1}{2001!} =

= \frac{1}{2!} - \frac{1}{2001!} = \frac{\frac{2001!}{2} - 1}{2001!} = \frac{2001! - 2}{2 * 2001!} ▴

* J a v o b * \frac{2001! - 2}{2 * 2001!}

A b d u l l a x a t o v J a s u r b e k

Commented by I want to learn more last updated on 06/Jun/20

Thanks sir .

Commented by 06122004 last updated on 06/Jun/20

o k!