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Algebra Level 4

1 1 4 + 1 2 + 1 + 2 2 4 + 2 2 + 1 + … + 2015 201 5 4 + 201 5 2 + 1 \large{ \dfrac{1}{1^4 + 1^2 + 1} + \dfrac{2}{2^4 + 2^2 + 1} + \ldots + \dfrac{2015}{2015^4 + 2015^2 + 1} }

If the value of the above can be expressed as A B \dfrac{A}{B} for positive coprime integers ( A , B ) (A,B) , submit the value of A + B A+B as your answer.

Inspiration


The answer is 6093361.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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3 solutions

This is a simple telescopic series.

G e n e r a l t e r m i s n n 4 + n 2 + 1 B y p a r t i a l f r a c t i o n n n 4 + n 2 + 1 = 1 2 { 1 n 2 − n + 1 − 1 n 2 + n + 1 } ∑ n = 1 2015 n n 4 + n 2 + 1 = 1 2 ∗ ∑ n = 1 2015 { 1 n 2 − n + 1 } − 1 2 ∗ ∑ n = 1 2015 { 1 n 2 + n + 1 } = 1 2 { 1 + ∑ n = 1 2014 { 1 n 2 + n + 1 } − ∑ n = 1 2014 { 1 n 2 + n + 1 } − 2015 201 5 2 + 2015 + 1 } = 1 2 { 1 − 1 201 5 2 + 2015 + 1 } = 1 2 { 1 − 1 4062241 } = 6093361 General ~term~ is~\dfrac n {n^4+n^2+1}\\ By~ partial~ fraction\\ \dfrac n {n^4+n^2+1}\\ =\dfrac 1 2 \left \{\dfrac 1 {n^2-n+1} ~-~\dfrac 1{n^2+n+1} \right \}\\ \displaystyle \sum_{n=1}^{2015} \dfrac n {n^4+n^2+1} \\ \displaystyle =\dfrac 1 2* \sum_{n=1}^{2015} \left \{\dfrac 1 {n^2-n+1} \right \}~-~\dfrac 1 2* \sum_{n=1}^{2015} \left \{\dfrac 1{n^2+n+1} \right \}\\ \displaystyle =\dfrac 1 2\left \{1+ \color{#3D99F6}{ \sum_{n=1}^{2014} \left \{\dfrac 1 {n^2+n+1} \right \}- \sum_{n=1}^{2014} \left \{\dfrac 1{n^2+n+1} \right \} }-\dfrac {2015}{2015^2+2015+1} \right \} \\= \dfrac 1 2\left \{1- \dfrac 1 {2015^2+ 2015+1} \right \} \\=\dfrac 1 2 \left \{1-\dfrac 1 {4062241} \right \} \\ =\Large \color{#D61F06}{6093361}

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Level 4 is too much for this sir !

Venkata Karthik Bandaru - 5 years, 9 months ago

You would not believe me I got my answer right in notebook right now but when I used Brilliant's Scratchpad I am getting the wrong answer. @Calvin Lin

Department 8 - 5 years, 9 months ago

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Thanks. It is a known bug that the scratchpad rounds off to 6 significant digits. We are working on fixing it.

Calvin Lin Staff - 5 years, 9 months ago

Can you elaborate the Partial Fractions

T h a n k y o u v e r y m u c h U p v o t e d \color{#D61F06}{Thank \quad you \quad very \quad much} \quad \color{grey}{Upvoted}

Syed Baqir - 5 years, 9 months ago

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n n 4 + n 2 + 1 = n ( n 2 + n + 1 ) ∗ ( n 2 − n + 1 ) = C n + A n 2 + n + 1 + D n + B n 2 − n + 1 ∴ n = ( C n + A ) ( n 2 − n + 1 ) + ( D n + B ) ( n 2 + n + 1 ) ∴ E q u a t i n g c o e f f i c e n t o f c o n s t a n t t e r m A + B = 0. ⟹ B = − A . . . . . ( ∗ ) ∴ E q u a t i n g c o e f f i c e n t o f n 3 t e r m C + D = 0.... ( ∗ ∗ ) ∴ E q u a t i n g c o e f f i c e n t o f t e r m n , 1 = − A + B + C + D . ⟹ 1 = − A − A b y ( ∗ ) a n d ( ∗ ∗ ) . ∴ A = − 1 2 , a n d B = 1 2 . ∴ E q u a t i n g c o e f f i c e n t o f n 2 t e r m , 0 = A + B − C + D = 0 − C + D . ( ∗ ∗ ∗ ) ⟹ C + D = 0 b y ( ∗ ∗ ) , − C + D = 0 b y ( ∗ ∗ ∗ ) , ⟹ C = D = 0. ⟹ n ( n 2 + n + 1 ) ∗ ( n 2 − n + 1 ) = 1 2 ∗ ( 1 n 2 − n + 1 − 1 n 2 + n + 1 ) . Sorry, I did not see Your comment earlier. \dfrac n {n^4+n^2+1}=\color{#D61F06}{\dfrac n {(n^2+n+1)*(n^2-n+1)} }\\ =\dfrac {Cn+A}{n^2+n+1} +\dfrac {Dn+B}{n^2-n+1 } \\ \therefore~ n= (Cn+A) (n^2-n+1)+ (Dn+B)(n^2+n+1) \\ \therefore ~Equating~coefficent~of~constant~term~ A+B=0.\\ \implies~B= - A.....(*)\\ \therefore ~Equating~coefficent~of~n^3~term~ C+D=0....(**)\\ \therefore~Equating~coefficent~of~term~n,~~~~~1=-A+B+C+D.\\ \implies~1= - A - A~~by~(*)\ ~and~(**).\\ \therefore~A=-\dfrac 1 2 ~~,~~and ~~B=\dfrac 1 2 .\\ \therefore ~Equating~coefficent~of~n^2~term ,~~~0=A+B-C+D =0-C+D.(***)\\ \implies~C+D=0~~by (**),~~-C+D=0~~by (***),~~\implies~C=D=0.\\ \implies~~\color{#D61F06}{\dfrac n {(n^2+n+1)*(n^2-n+1)} }\\ =\dfrac 1 2 * \left( \dfrac1{ n^2-n+1} - \dfrac 1 { n^2+n+1} \right ) . \\ \text{Sorry, I did not see Your comment earlier.}

Niranjan Khanderia - 5 years, 9 months ago

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T h a n k y o u ! ! ! \color{grey}{Thank \quad you \quad !!!}

Syed Baqir - 5 years, 9 months ago

Exactly Same way

Kushagra Sahni - 5 years, 9 months ago

Let the sum be S S ; then we have:

S = ∑ k = 1 2015 k k 4 + k 2 + 1 = ∑ k = 1 2015 k ( k 2 − 1 ) ( k 2 − 1 ) ( k 4 + k 2 + 1 ) = ∑ k = 1 2015 k ( k − 1 ) ( k + 1 ) k 6 − 1 = ∑ k = 1 2015 k ( k − 1 ) ( k + 1 ) ( k 3 + 1 ) ( k 3 − 1 ) = ∑ k = 1 2015 k ( k 2 − k + 1 ) ( k 2 + k + 1 ) = 1 2 ∑ k = 1 2015 ( 1 k 2 − k + 1 − 1 k 2 + k + 1 ) = 1 2 ∑ k = 1 2015 ( 1 ( k − 1 2 ) 2 + 3 4 − 1 ( k + 1 2 ) 2 + 3 4 ) = 1 2 ( ∑ k = 1 2015 1 ( k − 1 2 ) 2 + 3 4 − ∑ k = 2 2016 1 ( k − 1 2 ) 2 + 3 4 ) = 1 2 ( 1 ( 1 − 1 2 ) 2 + 3 4 − 1 ( 2016 − 1 2 ) 2 + 3 4 ) = 1 2 ( 1 − 1 ( 4031 2 ) 2 + 3 4 ) = 1 2 ( 1 − 1 4062241 ) = 2031120 4062241 \begin{aligned} S & = \sum_{k=1}^{2015} \frac{k}{k^4+k^2+1} = \sum_{k=1}^{2015} \frac{k(k^2-1)}{(k^2-1)(k^4+k^2+1)} = \sum_{k=1}^{2015} \frac{k(k-1)(k+1)}{k^6-1} \\ & = \sum_{k=1}^{2015} \frac{k(k-1)(k+1)}{(k^3+1)(k^3-1)} = \sum_{k=1}^{2015} \frac{k}{(k^2-k+1)(k^2+k+1)} \\ & = \frac{1}{2} \sum_{k=1}^{2015} \left(\frac{1}{k^2-k+1} - \frac{1}{k^2+k+1} \right) \\ & = \frac{1}{2} \sum_{k=1}^{2015} \left(\frac{1}{\left(k-\frac{1}{2}\right)^2+\frac{3}{4}} - \frac{1}{\left(k+\frac{1}{2}\right)^2+\frac{3}{4}} \right) \\ & = \frac{1}{2} \left( \sum_{k=1}^{2015} \frac{1}{\left(k-\frac{1}{2}\right)^2+\frac{3}{4}} - \sum_{k=2}^{2016} \frac{1}{\left(k-\frac{1}{2}\right)^2+\frac{3}{4}} \right) \\ & = \frac{1}{2} \left(\frac{1}{\left(1-\frac{1}{2}\right)^2+\frac{3}{4}} - \frac{1}{\left(2016-\frac{1}{2}\right)^2+\frac{3}{4}} \right) \\ & = \frac{1}{2} \left(1 - \frac{1}{\left(\frac{4031}{2}\right)^2+\frac{3}{4}} \right) = \frac{1}{2} \left(1 - \frac{1}{4062241} \right) = \frac{2031120}{4062241} \end{aligned}

Since 4062241 4062241 is a prime, 2031120 2031120 is its coprime.

Therefore, A + B = 2031120 + 4062241 = 6093361 A+B = 2031120 + 4062241 = \boxed{6093361}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

I too first did the same way but we can factor directly as under.

x 4 + x 2 + 1 = x 4 + 2 x 2 + 1 − x 2 = ( x 2 + 1 ) 2 − x 2 = . . . . . . x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=......

Niranjan Khanderia - 5 years, 9 months ago

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I see. Thanks.

Chew-Seong Cheong - 5 years, 9 months ago

U p v o t e d \color{grey}{Upvoted}

Syed Baqir - 5 years, 9 months ago
Karim Mohamed
Sep 3, 2015

the general form is :

∑ n = 1 2015 n n 4 + n 2 + 1 \sum _{ n=1 }^{ 2015 }{ \frac { n }{ { n }^{ 4 }+{ n }^{ 2 }+1 } }

the denominator can be factorized into :

n 4 + n 2 + 1 = ( n 2 + 1 ) 2 − n 2 = ( n 2 + n + 1 ) ( n 2 − n + 1 ) { n }^{ 4 }+{ n }^{ 2 }+1={ \left( { n }^{ 2 }+1 \right) }^{ 2 }-{ n }^{ 2 }=\left( { n }^{ 2 }+n+1 \right) \left( { n }^{ 2 }-n+1 \right)

by using partial fractions:

∑ n = 1 2015 n ( n 2 + n + 1 ) ( n 2 − n + 1 ) = − 1 2 ∑ n = 1 2015 ( 1 n 2 + n + 1 − 1 n 2 − n + 1 ) \sum _{ n=1 }^{ 2015 }{ \frac { n }{ \left( { n }^{ 2 }+n+1 \right) \left( { n }^{ 2 }-n+1 \right) } } =\frac { -1 }{ 2 } \sum _{ n=1 }^{ 2015 }{ \left( \frac { 1 }{ { n }^{ 2 }+n+1 } -\frac { 1 }{ { n }^{ 2 }-n+1 } \right) }

the first term can be manipulated as follows:

∑ n = 1 2015 1 n 2 + n + 1 = ∑ n = 1 2015 1 ( n + 1 ) 2 − n = ∑ m = 2 2016 1 m 2 − m + 1 \sum _{ n=1 }^{ 2015 }{ \frac { 1 }{ { n }^{ 2 }+n+1 } } =\sum _{ n=1 }^{ 2015 }{ \frac { 1 }{ { \left( n+1 \right) }^{ 2 }-n } } =\sum _{ m=2 }^{ 2016 }{ \frac { 1 }{ { m }^{ 2 }-m+1 } }

l e t m = n + 1 let\quad m=n+1

since m,n are arbitrary variables, use n for both:

− 1 2 ( ∑ n = 2 2016 1 n 2 − n + 1 − ∑ n = 1 2015 1 n 2 − n + 1 ) = \frac { -1 }{ 2 } \left( \sum _{ n=2 }^{ 2016 }{ \frac { 1 }{ { n }^{ 2 }-n+1 } } -\sum _{ n=1 }^{ 2015 }{ \frac { 1 }{ { n }^{ 2 }-n+1 } } \right) = − 1 2 ( 1 2016 2 − 2016 + 1 − 1 ) = \frac { -1 }{ 2 } \left( \frac { 1 }{ { 2016 }^{ 2 }-2016+1 } -1 \right) = − 1 2 ( 1 4062241 − 1 ) = 2031120 4062241 \frac { -1 }{ 2 } \left( \frac { 1 }{ 4062241 } -1 \right) =\frac { 2031120 }{ 4062241 }

note that at the previous equation all terms cancel each other except the terms at which n=2016, and n=1

so : A + B = 2031120 + 4062241 = 6093361 A+B=2031120+4062241=\boxed { 6093361 }

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