Simplify Each Of These Fractions First!

Algebra Level 3

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 \dfrac 1{1+1^2 + 1^4} + \dfrac 2{1+2^2 + 2^4} + \dfrac 3{1+3^2 + 3^4} +\cdots + \dfrac {99}{1+99^2 + 99^4}

The value of the expression above lies between which of the following pairs of numbers?

0.52 and 1 0.48 and 0.49 0.46 and 0.47 0.49 and 0.50 0.47 and 0.48

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Anshu Garg by anshu garg , 18, India

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2 solutions

S = ∑ n = 1 99 n n + n 2 + n 4 = ∑ n = 1 99 n ( n 2 − n + 1 ) ( n 2 + n + 1 ) = 1 2 ∑ n = 1 99 ( 1 n 2 − n + 1 − 1 n 2 + n + 1 ) = 1 2 ∑ n = 1 99 ( 1 n 2 − n + 1 − 1 ( n + 1 ) 2 − ( n + 1 ) + 1 ) = 1 2 ( ∑ n = 1 99 1 n 2 − n + 1 − ∑ n = 2 100 1 n 2 − n + 1 ) = 1 2 ( 1 1 − 1 10 0 2 − 100 + 1 ) > 1 2 ( 0.9999 ) = 0.49995 \begin{aligned} S & = \sum_{n=1}^{99} \frac n{n+n^2+n^4} \\ & = \sum_{n=1}^{99} \frac n{(n^2-n+1)(n^2+n+1)} \\ & = \frac 12 \sum_{n=1}^{99} \left(\frac 1{n^2-n+1} - \frac 1{n^2+n+1}\right) \\ & = \frac 12 \sum_{n=1}^{99} \left(\frac 1{n^2-n+1} - \frac 1{(n+1)^2-(n+1)+1}\right) \\ & = \frac 12 \left(\sum_{n=1}^{99} \frac 1{n^2-n+1} - \sum_{n=2}^{100} \frac 1{n^2-n+1} \right) \\ & = \frac 12 \left(\frac 11 - \frac 1{100^2-100+1} \right) \\ & > \frac 12 (0.9999) = 0.49995 \end{aligned}

⟹ 0.49 < S < 0.5 \implies \boxed{0.49 < S < 0.5} .

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Partial fractioning these rational functions often creates a telescoping which makes results transparent. Heaviside's method can compute numerators quickly, especially for unrepeated linear factors.

Will Heierman - 4 years, 7 months ago
Anshu Garg
Nov 6, 2016

Let us take: x 4 + x 2 + 1 = ( x 2 + x + 1 ) ( x 2 − x + 1 ) x^4+x^2+1=(x^2+x+1)(x^2-x+1)

Using this the above expression can be reduced into = 1 1.3 \frac{1}{1.3} + 2 3.7 \frac{2}{3.7} + 3 7.13 \frac{3}{7.13} ..................................................... 99 9703.9901 \frac{99}{9703.9901} )

Taking 1 2 \frac{1}{2} common from each term = 1 2 \frac{1}{2} [1- 1 3 \frac{1}{3} ]+ 1 2 \frac{1}{2} [ 1 3 \frac{1}{3} - 1 7 \frac{1}{7} ]+ 1 2 \frac{1}{2} [ 1 7 \frac{1}{7} - 1 13 \frac{1}{13} ]................................................ 1 2 \frac{1}{2} [ 1 9703 \frac{1}{9703} - 1 9901 \frac{1}{9901} ]

= 1 2 \frac{1}{2} [1- 1 3 \frac{1}{3} + 1 3 \frac{1}{3} - 1 7 \frac{1}{7} +.................................................... 1 9703 \frac{1}{9703} - 1 9901 \frac{1}{9901} ] = 4500 9901 \frac{4500}{9901}

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