Factorial sum!

Algebra Level 4

ζ = ∑ r = 1 2016 r 2 − r − 1 ( r + 1 ) ! \large \zeta = \sum_{r=1}^{2016} \dfrac{r^2-r-1}{(r+1)!}

If ζ \zeta can be expressed as − 1 a ( b ! ) -\dfrac1{a(b!)} , where a a and b b are positive integers, find a + b a+b .

Notation :

! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 4032.

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Aditya Dhawan by Aditya Dhawan , 20, India

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2 solutions

Rishabh Jain
May 1, 2016

ζ = ∑ r = 1 2016 ( r 2 − 1 ⏞ ( r − 1 ) ( r + 1 ) − r ( r + 1 ) ! ) = ∑ r = 1 2016 ( ( r − 1 ) ( r + 1 ) ( r + 1 ) r ! − r ( r + 1 ) ! ) = ∑ r = 1 2016 ( r − 1 r ! − r ( r + 1 ) ! ) ( A T e l e s c o p i c S e r i e s ) = 0 − 2016 2017 ! = − 1 2017 ( 2015 ! ) \large{\begin{aligned}\zeta=&\displaystyle\sum_{r=1}^{2016}\left(\dfrac{\overbrace{r^2-1}^{(r-1)(r+1)}-r}{(r+1)!}\right)\\=& \displaystyle\sum_{r=1}^{2016}\left(\dfrac{(r-1)\cancel{(r+1)}}{\cancel{(r+1)}r!}-\dfrac{r}{(r+1)!}\right)\\=& \displaystyle\sum_{r=1}^{2016}\left(\dfrac{r-1}{r!}-\dfrac{r}{(r+1)!}\right)\\&\\&\mathbf{(A~Telescopic~Series)}\\&\\=&0-\dfrac{2016}{2017!}=\dfrac{-1}{2017(2015!)}\end{aligned}}

2015 + 2017 = 4032 \Large 2015+2017=\boxed{\huge 4032}

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A simple yet elegant solution! (+1)

Aditya Dhawan - 5 years, 1 month ago

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Thanks.... ⌣ ¨ \ddot\smile

Rishabh Jain - 5 years, 1 month ago
Aditya Dhawan
May 1, 2016

ζ = ∑ r = 1 2016 r 2 − r − 1 ( r + 1 ) ! = ∑ r = 1 n r ( r + 1 ) − 2 r − 1 ( r + 1 ) ! = = ∑ r = 1 n 1 ( r − 1 ) ! − 2 r ( r + 1 ) ! − 1 ( r + 1 ) ! = ∑ r = 1 n 1 ( r − 1 ) ! − 1 ( r + 1 ) ! − 2 ∑ r = 1 n r ( r + 1 ) ! N o w , ∑ r = 1 n r ( r + 1 ) ! = ∑ ( r + 1 ) − 1 ( r + 1 ) ! = ∑ r = 1 n 1 r ! − 1 ( r + 1 ) ! = 1 − 1 ( n + 1 ) ! ( T e l e s c o p i n g S e r i e s ) S i m i l a r l y , ∑ r = 1 n 1 ( r − 1 ) ! − 1 ( r + 1 ) ! = 2 − 1 n ! − 1 ( n + 1 ) ! T h u s t h e c l o s e d f o r m i s : − 1 n ! + 1 ( n + 1 ) ! = − n ( n + 1 ) ! = − 1 ( n − 1 ) ( n + 1 ) ! ∴ ζ = − 1 ( 2015 ) ( 2017 ! ) ⟹ a = 2015 , b = 2017 ∴ a + b = 4032 \zeta =\sum _{ r=1 }^{ 2016 }{ \frac { { r }^{ 2 }-r-1 }{ (r+1)! } } \\ =\sum _{ r=1 }^{ n }{ \frac { r(r+1)-2r-1 }{ (r+1)! } } =\\ =\sum _{ r=1 }^{ n }{ \frac { 1 }{ (r-1)! } -\frac { 2r }{ (r+1)! } -\frac { 1 }{ (r+1)! } } \\ =\sum _{ r=1 }^{ n }{ \frac { 1 }{ (r-1)! } -\frac { 1 }{ (r+1)! } } -\quad 2\sum _{ r=1 }^{ n }{ \frac { r }{ (r+1)! } } \\ Now,\\ \sum _{ r=1 }^{ n }{ \frac { r }{ (r+1)! } } =\sum { \frac { (r+1)-1 }{ (r+1)! } } =\sum _{ r=1 }^{ n }{ \frac { 1 }{ r! } -\frac { 1 }{ (r+1)! } } =1-\frac { 1 }{ (n+1)! } \left( Telescoping\quad Series \right) \\ Similarly,\quad \\ \sum _{ r=1 }^{ n }{ \frac { 1 }{ (r-1)! } -\frac { 1 }{ (r+1)! } } =\quad 2-\frac { 1 }{ n! } -\frac { 1 }{ (n+1)! } \\ Thus\quad the\quad closed\quad form\quad is:\quad -\frac { 1 }{ n! } +\frac { 1 }{ (n+1)! } =\frac { -n }{ (n+1)! } =\frac { -1 }{ (n-1)(n+1)! } \\ \therefore \quad \zeta =\frac { -1 }{ (2015)(2017!) } \Longrightarrow \quad a=2015,b=2017\\ \therefore \quad \boxed { a+b=4032 } \quad \\

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Moderator note:

Good recognition of the telescoping series.

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