Sequences and series (8)

Algebra Level 4

The n t h n^{th} term of the sequence is given by t n = n 5 + n 3 n 4 + n 2 + 1 t_{n}= \dfrac{n^5 +n^3}{n^4+n^2+1} and if sum of its n n terms can be expressed as S n = a n 2 + a + 1 b n 2 + b S_{n}=a_{n}^2 +a+ \dfrac{1}{b_{n}^2 +b} , where a n a_{n} and b n b_{n} are the n t h n^{th} terms of some arithmetic progression and a , b a,b are some constants. Then find the positive integral value of b n a n \dfrac{b_{n}}{a_{n}} .


The answer is 2.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

We note that:

S n = ∑ k = 1 n k 5 + k 3 k 4 + k 2 + 1 = ∑ k = 1 n k ( k 4 + k 2 + 1 ) − k k 4 + k 2 + 1 = ∑ k = 1 n ( k − k k 4 + k 2 + 1 ) = ∑ k = 1 n ( k − 1 2 ( 1 k 2 − k + 1 − 1 k 2 + k + 1 ) ) = ∑ k = 1 n k − 1 2 ( ∑ k = 1 n 1 k 2 − k + 1 − ∑ k = 1 n 1 k 2 + k + 1 ) = ∑ k = 1 n k − 1 2 ( ∑ k = 1 n 1 k 2 − k + 1 − ∑ k = 2 n + 1 1 k 2 − k + 1 ) = n ( n + 1 ) 2 − 1 2 ( 1 − 1 n 2 + n + 1 ) = n 2 + n 2 − 1 2 + 1 2 ( n 2 + n + 1 ) Note that n 2 + n = ( 3 2 ⏟ = a + ( n − 1 ) ( 1 ) ⏟ = d ) 2 − 1 4 = ( 3 2 2 + ( n − 1 ) 1 2 ) 2 − 5 8 + 1 ( 3 2 2 + ( n − 1 ) 2 ) 2 + 1 2 \begin{aligned} S_n & = \sum_{k=1}^n \frac {k^5+k^3}{k^4+k^2+1} \\ & = \sum_{k=1}^n \frac {k(k^4+k^2+1)-k}{k^4+k^2+1} \\ & = \sum_{k=1}^n \left( k - \frac k{k^4+k^2+1} \right) \\ & = \sum_{k=1}^n \left( k - \frac 12 \left( \frac 1{k^2-k+1} - \frac 1{k^2+k+1} \right) \right) \\ & = \sum_{k=1}^n k - \frac 12 \left( \sum_{k=1}^n \frac 1{k^2-k+1} - \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}n} \frac 1{\color{#3D99F6}k^2+k+1} \right) \\ & = \sum_{k=1}^n k - \frac 12 \left( \sum_{k=1}^n \frac 1{k^2-k+1} - \sum_{\color{#D61F06}k=2}^{\color{#D61F06}n+1} \frac 1{\color{#D61F06}k^2-k+1} \right) \\ & = \frac {n(n+1)}2 - \frac 12 \left(1 - \frac 1{n^2+n+1} \right) \\ & = \frac {\color{#3D99F6}n^2+n}2 - \frac 12 + \frac 1{2({\color{#3D99F6}n^2+n}+1)} & \small \color{#3D99F6} \text{Note that } n^2 + n = \big(\underbrace{\frac 32}_{=a} + (n-1)\underbrace{(1)}_{=d} \big)^2 - \frac 14 \\ & = \left({\color{#3D99F6}\frac 3{2\sqrt 2}} + (n-1){\color{#3D99F6}\frac 1{\sqrt 2}} \right)^2 - \frac 58 + \frac 1{ \left({\color{#3D99F6}\frac {3\sqrt 2}2} + (n-1){\color{#3D99F6}\sqrt 2} \right)^2+\frac 12} \end{aligned}

⟹ b n a n = 3 2 2 + ( n − 1 ) 2 3 2 2 + ( n − 1 ) 1 2 = 2 \implies \dfrac {b_n}{a_n} = \dfrac {\frac {3\sqrt 2}2 + (n-1)\sqrt 2}{\frac 3{2\sqrt 2} + (n-1)\frac 1{\sqrt 2}} = \boxed{2}

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