one of NTSE's best # 1

Algebra Level 3

x = 1 1 + 1 2 + 1 4 + 2 1 + 2 2 + 2 4 + 3 1 + 3 2 + 3 4 + ⋯ 99 1 + 9 9 2 + 9 9 4 x = \frac 1{1+1^2+1^4} + \frac 2{1+2^2+2^4} + \frac 3{1+3^2+3^4} + \cdots \frac {99}{1+99^2+99^4}

What values does x x lie in between?

0.48 < x < 0.49 0.48<x<0.49 0.47 < x < 0.48 0.47<x<0.48 0.49 < x < 0.50 0.49< x <0.50 0.46 < x < 0.47 0.46<x<0.47

Suresh Jh by suresh jh , 17, India

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2 solutions

Chew-Seong Cheong
Feb 11, 2018

x = ∑ k = 1 n k 1 + k 2 + k 4 where n = 99 = ∑ k = 1 n k ( k 2 − k + 1 ) ( k 2 + k + 1 ) Note that k 4 + k 2 + 1 = ( k 2 − k + 1 ) ( k 2 + k + 1 ) = 1 2 ∑ k = 1 n ( 1 k 2 − k + 1 − 1 k 2 + k + 1 ) Note that k 2 + k + 1 = ( k + 1 ) 2 − ( k + 1 ) + 1 = 1 2 ( ∑ k = 1 n 1 k 2 − k + 1 − ∑ k = 1 n 1 ( k + 1 ) 2 − ( k + 1 ) + 1 ) Let j = k + 1 = 1 2 ( ∑ k = 1 n 1 k 2 − k + 1 − ∑ j = 2 n + 1 1 j 2 − j + 1 ) Replace j with k = 1 2 ( ∑ k = 1 n 1 k 2 − k + 1 − ∑ k = 2 n + 1 1 k 2 − k + 1 ) = 1 2 ( 1 1 2 − 1 + 1 − 1 ( n + 1 ) 2 − ( n + 1 ) + 1 ) Putting back n = 99 = 1 2 ( 1 − 1 10 0 2 − 100 + 1 ) \begin{aligned} x & = \sum_{k=1}^{\color{#D61F06}n} \frac k{\color{#3D99F6}1+k^2+k^4} & \small \color{#D61F06} \text{where }n = 99 \\ & = \sum_{k=1}^n \frac k{\color{#3D99F6}(k^2-k+1)(k^2+k+1)} & \small \color{#3D99F6} \text{Note that }k^4+k^2+1 = (k^2-k+1)(k^2+k+1) \\ & = \frac 12 \sum_{k=1}^n \left(\frac 1{k^2-k+1} - \frac 1{\color{#3D99F6}k^2+k+1}\right) & \small \color{#3D99F6} \text{Note that }k^2+k+1 = (k+1)^2-(k+1)+1 \\ & = \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+1)^2-(k+1)+1} \right) & \small \color{#3D99F6} \text{Let }j = k+1 \\ & = \frac 12 \left( \sum_{k=1}^n \frac 1{k^2-k+1} - \sum_{\color{#3D99F6}j=2}^{\color{#3D99F6}n+1} \frac 1{\color{#3D99F6}j^2-j+1} \right) & \small \color{#3D99F6} \text{Replace }j \text{ with }k \\ & = \frac 12 \left( \sum_{k=1}^n \frac 1{k^2-k+1} - \sum_{\color{#3D99F6}k=2}^{\color{#3D99F6}n+1} \frac 1{\color{#3D99F6}k^2-k+1} \right) \\ & = \frac 12 \left(\frac 1{1^2-1+1} - \frac 1{(n+1)^2-(n+1)+1} \right) & \small \color{#3D99F6} \text{Putting back }n=99 \\ & = \frac 12 \left(1 - \frac 1{100^2-100+1} \right) \end{aligned}

Therefore, 1 2 − 1 100 < x < 1 2 ⟹ 0.49 < x < 0.50 \frac 12 - \frac 1{100}< x < \frac 12 \implies \boxed{0.49 < x < 0.50}

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I liked the coloring of this solution. Nice sir.

Vilakshan Gupta - 3 years, 3 months ago

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Thanks. Glad that you like it. No upvote?

Chew-Seong Cheong - 3 years, 3 months ago

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Upvoted ! Initially i didn't remember to upvote it.

Vilakshan Gupta - 3 years, 3 months ago

Sir, in the fourth last step, could you explain the red coloured where you changed the bounds from k=2 to n+1 .

Aman thegreat - 3 years, 3 months ago

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I have added a line to explained.

Chew-Seong Cheong - 3 years, 3 months ago
Suresh Jh
Feb 10, 2018

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