'Bashing' Telescope.

Algebra Level 5

∑ i = 2 10 4 i 5 + 10 i 4 + 16 i 3 + 14 i 2 − 4 i − 4 i 8 + 4 i 7 + 2 i 6 − 8 i 5 − 7 i 4 + 4 i 3 + 4 i 2 \large \displaystyle \sum_{i=2}^{10} \dfrac{4i^5 + 10i^4 + 16i^3 + 14i^2-4i - 4}{i^8 + 4i^7 + 2i^6 - 8i^5 - 7i^4 + 4i^3 + 4i^2}

If the above summation can be simplified to a b \dfrac{a}{b} where a a and b b are coprime positive integers . What is the value of a + b a+b ?


The answer is 2513.

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Nihar Mahajan by Nihar Mahajan , 20, India

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1 solution

Nihar Mahajan
Mar 8, 2015

∑ i = 2 10 4 i 5 + 10 i 4 + 16 i 3 + 14 i 2 − 4 i − 4 i 8 + 4 i 7 + 2 i 6 − 8 i 5 − 7 i 4 + 4 i 3 + 4 i 2 \displaystyle \sum_{i=2}^{10} \dfrac{4i^5 + 10i^4 + 16i^3 + 14i^2-4i - 4}{i^8 + 4i^7 + 2i^6 - 8i^5 - 7i^4 + 4i^3 + 4i^2}

It can be written as:

∑ i = 2 10 i 6 + 6 i 5 + 13 i 4 + 12 i 3 + 4 i 2 − ( i 6 + 4 i 5 + 2 i 4 − 8 i 3 − 7 i 2 + 4 i + 4 ) + ( i 6 + 2 i 5 − 3 i 4 − 4 i 3 + 4 i 2 − ( i 6 − 2 i 4 + i 2 ) ( i − 1 ) 2 ( i ) 2 ( i + 1 ) 2 ( i + 2 ) 2 \displaystyle \sum_{i=2}^{10} \dfrac{i^6 + 6i^5 + 13i^4 + 12i^3+4i^2 \\ -(i^6 + 4i^5 + 2i^4 - 8i^3 - 7i^2 + 4i + 4 ) \\+(i^6 + 2i^5 - 3i^4 - 4i^3 + 4i^2 \\ - (i^6 - 2i^4 + i^2) }{(i-1)^2(i)^2(i+1)^2(i+2)^2}

After factorizing:

∑ i = 2 10 ( i ) 2 ( i + 1 ) 2 ( i + 2 ) 2 − ( i − 1 ) 2 ( i + 1 ) 2 ( i + 2 ) 2 + ( i − 1 ) 2 ( i ) 2 ( i + 2 ) 2 − ( i − 1 ) 2 ( i ) 2 ( i + 1 ) 2 ( i − 1 ) 2 ( i ) 2 ( i + 1 ) 2 ( i + 2 ) 2 \displaystyle \sum_{i=2}^{10} \dfrac{(i)^2(i+1)^2(i+2)^2 \\- (i-1)^2(i+1)^2(i+2)^2 \\+ (i-1)^2(i)^2(i+2)^2 \\- (i-1)^2(i)^2(i+1)^2}{(i-1)^2(i)^2(i+1)^2(i+2)^2}

∑ i = 2 10 1 ( i − 1 ) 2 − 1 i 2 + 1 ( i + 1 ) 2 − 1 ( i + 2 ) 2 \displaystyle \sum_{i=2}^{10} \dfrac{1}{(i-1)^2} - \dfrac{1}{i^2} + \dfrac{1}{(i+1)^2} - \dfrac{1}{(i+2)^2}

S = 1 1 − 1 4 + 1 9 − 1 16 1 4 − 1 9 + 1 16 − 1 25 … … … … 1 64 − 1 81 + 1 100 − 1 121 1 81 − 1 100 + 1 121 − 1 144 S=\dfrac{1}{1} - \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{1}{16} \\ \dfrac{1}{4} - \dfrac{1}{9} + \dfrac{1}{16} - \dfrac{1}{25} \\ \dots\\\dots\\\dots\\\dots \\ \dfrac{1}{64} - \dfrac{1}{81} + \dfrac{1}{100} - \dfrac{1}{121} \\ \dfrac{1}{81} - \dfrac{1}{100} + \dfrac{1}{121} - \dfrac{1}{144}

S = 1 1 + 1 9 − 1 100 − 1 144 S = \dfrac{1}{1} + \dfrac{1}{9} - \dfrac{1}{100} - \dfrac{1}{144}

After simplification ,

S = 1313 1200 S = \dfrac{1313}{1200}

a + b = 1313 + 1200 = 2513 \huge a + b = 1313 + 1200 = \color{#D61F06}{\boxed{2513}}

31 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow I finally solved it.

After so many years cries

Mehul Arora - 5 years, 3 months ago

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Me too bro. Can't stop crying ! :))(

Aditya Sky - 5 years ago

NICE QUESTION WITH AN EVEN BETTER SOLUTION

Vaibhav Prasad - 6 years, 2 months ago

@Nihar Mahajan ,+1 ;D

Parth Lohomi - 6 years, 3 months ago

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:) :) try this 300 followers problem

Nihar Mahajan - 6 years, 3 months ago

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