Series Demystified 5

Algebra Level 5

∑ n = 1 46 ( n 3 − n 2 − 3 n − 4 ) ( n − 1 ) ! \large \sum_{n=1}^{46} \frac {(n^3-n^2-3n-4)}{(n-1)!}

If the summation above equals to S S , evaluate ( 46 ! S − 1 ) 1 3 (46!S-1)^{\frac{1}{3}} .

This is one of my original Series Demystified problems .


The answer is -47.

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Sanjeet Raria by Sanjeet Raria , 27, India

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3 solutions

Ayush Verma
Jun 26, 2015

We will try to change the expression in a telescoping sum.So let,

n 3 − n 2 − 3 n − 4 = A { ( n − 1 ) ( n − 2 ) ( n − 3 ) − ( n − 1 ) ( n − 2 ) } + B { ( n − 1 ) ( n − 2 ) − ( n − 1 ) } + C { ( n − 1 ) − 1 } c o m p a r e c o e f f i c i e n t o f n 3 ⇒ A = 1 p u t n = 1 ⇒ − C = − 7 o r C = 7 p u t n = 2 ⇒ − B = − 6 o r B = 6 ∴ ∑ n = 1 46 n 3 − n 2 − 3 n − 4 ( n − 1 ) ! = ∑ n = 1 46 { 1 ( n − 4 ) ! − 1 ( n − 3 ) ! } + 6 ∑ n = 1 46 { 1 ( n − 3 ) ! − 1 ( n − 2 ) ! } + 7 ∑ n = 1 46 { 1 ( n − 2 ) ! − 1 ( n − 1 ) ! } r e m e m b e r , 1 ( − m ) ! = 0 i f m ∈ I + , n o w S = − 1 43 ! − 6 44 ! − 7 45 ! = − 44 × 45 × 46 − 6 × 45 × 46 − 7 × 46 46 ! = − 103822 46 ! ⇒ ( 46 ! S − 1 ) 1 3 = ( − 103823 ) 1 3 = − 47 { n }^{ 3 }-{ n }^{ 2 }-3n-4=A\left\{ \left( n-1 \right) \left( n-2 \right) \left( n-3 \right) -\left( n-1 \right) \left( n-2 \right) \right\} \\ \\ +B\left\{ \left( n-1 \right) \left( n-2 \right) -\left( n-1 \right) \right\} +C\left\{ \left( n-1 \right) -1 \right\} \\ \\ compare\quad coefficient\quad of\quad { n }^{ 3 }\Rightarrow A=1\\ \\ put\quad n=1\Rightarrow -C=-7\quad or\quad C=7\\ \\ put\quad n=2\Rightarrow -B=-6\quad or\quad B=6\\ \\ \therefore \sum _{ n=1 }^{ 46 }{ \cfrac { { n }^{ 3 }-{ n }^{ 2 }-3n-4 }{ \left( n-1 \right) ! } } \\ \\ =\sum _{ n=1 }^{ 46 }{ \left\{ \cfrac { 1 }{ \left( n-4 \right) ! } -\cfrac { 1 }{ \left( n-3 \right) ! } \right\} +6\sum _{ n=1 }^{ 46 }{ \left\{ \cfrac { 1 }{ \left( n-3 \right) ! } -\cfrac { 1 }{ \left( n-2 \right) ! } \right\} } } \\ \\ +7\sum _{ n=1 }^{ 46 }{ \left\{ \cfrac { 1 }{ \left( n-2 \right) ! } -\cfrac { 1 }{ \left( n-1 \right) ! } \right\} } \\ \\ remember,\cfrac { 1 }{ \left( -m \right) ! } =0\quad if\quad m\in { I }^{ + },now\\ \\ S=-\cfrac { 1 }{ 43! } -\cfrac { 6 }{ 44! } -\cfrac { 7 }{ 45! } \\ \\ =\cfrac { -44\times 45\times 46-6\times 45\times 46-7\times 46 }{ 46! } =\cfrac { -103822 }{ 46! } \\ \\ \Rightarrow { \left( 46!S-1 \right) }^{ \cfrac { 1 }{ 3 } }={ \left( -103823 \right) }^{ \cfrac { 1 }{ 3 } }=-47

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Moderator note:

Simple standard approach.

Sourabh Jangid
Dec 1, 2016

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@Sourabh Jangid How did you think if such a split?

Ankit Kumar Jain - 4 years, 3 months ago

∴ T o a v o i d f a c t o r i a l s o f n e g a t i v e i n t e g e r : − S = S 1 − 4 + S 5 − 46 = ∑ n = 1 4 n 3 − n 2 − 3 n − 4 ( n − 1 ) ! + ∑ n = 5 46 n 3 − n 2 − 3 n − 4 ( n − 1 ) ! S 1 − 4 = ∑ n = 1 4 n 3 − n 2 − 3 n − 4 ( n − 1 ) ! = − 7 1 + − 6 1 + 5 2 + 32 6 = − 5 1 6 . n 3 − n 2 − 3 n − 4 = ( n − 3 ) ( n − 2 ) ( n − 1 ) + 5 ( n − 2 ) ( n − 1 ) + ( n − 1 ) − 7. S 5 − 46 = ∑ n = 5 46 n 3 − n 2 − 3 n − 4 ( n − 1 ) ! = ∑ n = 5 46 ( n − 3 ) ( n − 2 ) ( n − 1 ) + 5 ( n − 2 ) ( n − 1 ) + ( n − 1 ) − 7 ( n − 1 ) ! = ∑ n = 5 46 { 1 ( n − 4 ) ! + 5 ( n − 3 ) ! + 1 ( n − 2 ) ! − 7 ( n − 1 ) ! } = ∑ n = 5 46 { 1 ( n − 4 ) ! − 1 ( n − 3 ) ! + 6 ( n − 3 ) ! − 6 ( n − 2 ) ! + 7 ( n − 2 ) ! − 7 ( n − 1 ) ! } = { 1 1 ! − 1 43 ! + 6 2 ! − 6 44 ! + 7 3 ! − 7 45 ! } = 1 − 1 43 ! + 3 − 6 44 ! + 1 1 6 − 7 45 ! = + 5 1 6 − 1 43 ! − 6 44 ! − 7 45 ! . S = S 1 − 4 + S 5 − 46 = − 5 1 6 + 5 1 6 − 1 43 ! − 6 44 ! − 7 45 ! = − 1 43 ! − 6 44 ! − 7 45 ! . ⟹ S ∗ 46 ! − 1 = 46 ! 43 ! + 6 ∗ 46 ! 44 ! + 7 ∗ 46 ! 45 ! − 1 = − 44 ∗ 45 ∗ 46 − 6 ∗ 45 ∗ 46 − 7 ∗ 46 − 1 = − 103823. ∴ − 103823 3 = − 47 \color{#3D99F6}{ \\ \displaystyle \therefore~To~avoid~factorials~of~negative~integer:-~~~~S=S_{1-4}~+~S_{5-46}=~\sum_{n=1}^{4}\dfrac{n^3-n^2-3n-4}{(n-1)!}+\sum_{n=5}^{46}\dfrac{n^3-n^2-3n-4}{(n-1)!} \\ S_{1-4} \displaystyle =\sum_{n=1}^4\dfrac{n^3-n^2-3n-4}{(n-1)!}=\dfrac {-7} 1+\dfrac {-6} 1+\dfrac 5 2+\dfrac {32} 6= - 5\frac16.\\ n^3-n^2-3n-4 = (n-3)(n-2)(n-1)+5(n-2)(n-1)+(n-1) -7 .\\ \displaystyle S_{5-46}=\sum_{n=5}^{46}\dfrac{n^3-n^2-3n-4}{(n-1)!}=\sum_{n=5}^{46}\dfrac{(n-3)(n-2)(n-1)+5(n-2)(n-1)+(n-1) - 7}{(n-1)!} \\ \displaystyle =\sum_{n=5}^{46} \Big \{\dfrac 1 {(n-4)!} +\dfrac 5 {(n-3)!}+ \dfrac 1 {(n-2)!} -\dfrac 7 {(n-1)!} \Big\}\\ \displaystyle =\sum_ {n=5}^{46} \Big \{\dfrac 1 {(n-4)!} - \dfrac 1 {(n-3)!}~+~\dfrac 6 {(n-3)!} - \dfrac 6 {(n-2)!}~+~\dfrac 7 {(n-2)!} - \dfrac 7 {(n-1)!} \Big \} \\ \displaystyle = \Big \{\dfrac 1 {1!} - \dfrac 1 {43!}~+~\dfrac 6 {2!} - \dfrac 6 {44!}~+~\dfrac 7 {3!} - \dfrac 7 {45!} \Big \} \\ =1~-~\dfrac 1 {43!}~+~3~-~\dfrac 6 {44!} ~+~1\frac1 6 - \dfrac 7 {45!}=+5\frac1 6 - \dfrac 1 {43!}~-~\dfrac 6 {44!} ~- \dfrac 7 {45!}.\\ S=S_{1-4}+S_{5-46}= -5\frac1 6+5\frac1 6-\dfrac 1{43!}~-~\dfrac 6 {44!} - \dfrac 7 {45!}= - \dfrac 1 {43!}~-~\dfrac 6 {44!} ~- \dfrac 7 {45!}.\\ \implies~S*46!-1= \dfrac{46!} {43!}~+~\dfrac { 6*46!} {44!} ~+ \dfrac {7*46!} {45!}-1\\ = - 44*45*46 - 6*45*46 - 7*46 - 1= - 103823.\\ \therefore\sqrt[3]{ - 103823} =\Large~~~\color{#D61F06}{ - 47} } \\

I f w e a s s u m e t h a t f a c t o r i a l o f r e c i p r o c a l o f n e g a t i v e i n t e g e r i s 0 , 1 ( − n ) ! = 0 , t h e n , S = ∑ n = 1 46 n 3 − n 2 − 3 n − 4 ( n − 1 ) ! = ∑ n = 1 46 ( n − 3 ) ( n − 2 ) ( n − 1 ) + 5 ( n − 2 ) ( n − 1 ) + ( n − 1 ) − 7 ( n − 1 ) ! = ∑ n = 1 46 { 1 ( n − 4 ) ! + 5 ( n − 3 ) ! + 1 ( n − 2 ) ! − 7 ( n − 1 ) ! } ∑ n = 1 46 { 1 ( n − 4 ) ! − 1 ( n − 3 ) ! + 6 ( n − 3 ) ! − 6 ( n − 2 ) ! + 7 ( n − 2 ) ! − 7 ( n − 1 ) ! } = 0 − 1 43 ! + 0 − 6 44 ! + 0 − 7 45 ! ) ⟹ S ∗ 46 ! − 1 = − 46 ! 43 ! − 6 ∗ 46 ! 44 ! − 7 ∗ 46 ! 45 ! − 1 = − 44 ∗ 45 ∗ 46 − 6 ∗ 45 ∗ 46 − 7 ∗ 46 − 1 = − 103823. ∴ − 103823 3 = − 47 . \color{#BA33D6}{\\ If~we~assume~that~factorial~of~reciprocal~of~negative~integer~is~0, \dfrac 1{(-n)!}=0,~~then,\\ \displaystyle S=\sum_{n=1}^{46}\dfrac{n^3-n^2-3n-4}{(n-1)!}= \sum_{n=1}^{46}\dfrac{(n-3)(n-2)(n-1)+5(n-2)(n-1)+(n-1) - 7}{(n-1)!} \\ \displaystyle =\sum_{n=1}^{46} \Big \{\dfrac 1 {(n-4)!} +\dfrac 5 {(n-3)!}+ \dfrac 1 {(n-2)!} -\dfrac 7 {(n-1)!} \Big\}\\ \displaystyle \sum_ {n=1}^{46} \Big \{\dfrac 1 {(n-4)!} - \dfrac 1 {(n-3)!}~+~\dfrac 6 {(n-3)!} - \dfrac 6 {(n-2)!}~+~\dfrac 7 {(n-2)!} - \dfrac 7 {(n-1)!} \Big \} \\ =0 ~-~\dfrac 1 {43!}~+~0 ~-~\dfrac 6 {44!} ~+0 ~-~ \dfrac 7 {45!} )\\ \implies~S*46!-1= - \dfrac{46!} {43!}~-~\dfrac { 6*46!} {44!} ~- \dfrac {7*46!} {45!}-1\\ = - 44*45*46 - 6*45*46 - 7*46 - 1= - 103823.\\ \therefore\sqrt[3]{ - 103823} =\Large~~~\color{#D61F06}{ - 47}. } \\

I f o u n d n 3 − n 2 − 3 n − 4 = ( n − 3 ) ( n − 2 ) ( n − 1 ) + 5 ( n − 2 ) ( n − 1 ) + ( n − 1 ) − 7 , a s f o l l o w s . I t b e c o m e s s i m p l e i f t h e n u m e r a t o r i s a s i m p l e c o n s t a n t . D e n o m i n a t o r i s a f a c t o r i a l a n d n u m e r a t o r a c u b i c , s u g g e s t s w e c o u l d r e d u c e d e n o m i n a t o r b y t h r e e t e r m s , ( n − 3 ) ( n − 2 ) ( n − 1 ) , t o e l i m i n a t e t h e c u b i c t e r m i n t h e n u m e r a t o r . S i m i l a r l y r e d u c e t w o t e r m s i n d e n o m i n a t o r a n d e l i m i n a t e t h e q u a d r a t i c t e r m , . ( n − 2 ) ( n − 1 ) i n t h e n u m e r a t o r . A n d s o o n . T o e l i m i n a t e n 3 w e u s e t h e s e t h r e e t e r m s . ( n − 3 ) ( n − 2 ) ( n − 1 ) = n 3 − 6 n 2 + 11 n − 6. B u t w e h a v e n 3 − n 2 − 3 n − 4. S o w e a r e l e f t w i t h ( n 3 − n 2 − 3 n − 4 ) − ( n 3 − 6 n 2 + 11 n − 6 ) = 5 n 2 − 14 n + 2. T o e l i m i n a t e 5 ‾ n 2 w e u s e t h e s e t w o t e r m s 5 ( n − 2 ) ( n − 1 ) = 5 n 2 − 15 n + 10. B u t w e h a v e 5 n 2 − 14 n + 2. S o w e a r e l e f t w i t h ( 5 n 2 − 14 n + 2 ) − ( 5 n 2 − 15 n + 10 ) = n − 8. T o e l i m i n a t e 1 ‾ n w e u s e t h i s o n e t e r m ( n − 1 ) . B u t w e h a v e n − 8. S o w e a r e l e f t w i t h ( n − 8 ) − ( n − 1 ) = − 7 \color{#20A900}{ \\ I~found~~~n^3-n^2-3n-4 = (n-3)(n-2)(n-1)+5(n-2)(n-1)+(n-1)-7,~~as~follows.\\ It~becomes~simple~if~the~numerator~is~a~simple~constant.~~Denominator~is~a~factorial~and\\ numerator~a~cubic,~suggests~we~could~reduce~denominator ~ by~three~terms,~~(n-3)(n-2)(n-1),~to~eliminate~ \\ the~cubic~term~in~the~numerator.~~Similarly~reduce~two~terms~in~denominator~and~eliminate\\ the~quadratic~term,~~.(n-2)(n-1)~in~the~numerator.~And~so~on.\\ To~eliminate~n^3~we~use~these~three~terms~.{\color{#3D99F6}{ (n-3)(n-2)(n-1)}}=n^3 - 6n^2+11n-6.~But~we~have~n^3-n^2-3n-4. \\ So~we~are~left~with~(n^3-n^2-3n-4)-(n^3 - 6n^2+11n-6)=5n^2-14n+2.\\ To~eliminate~\overline{5}n^2~we~use~these~two~terms~~{\color{#3D99F6}{5(n-2)(n-1)}}=5n^2-15n+10.~~But~we~have~5n^2-14n+2. \\ So~we~are~left~with~(5n^2-14n+2) - (5n^2-15n+10)=n - 8.\\ To~eliminate~\overline{1}n~we~use~this~one~term~{\color{#3D99F6}{(n-1)}}.~But~we~have~n - 8.\\ So~we~are~left~with~(n - 8) -(n - 1)={\color{#3D99F6}{-7}} }

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