Welcome 2016! Part 21

Algebra Level 3

5 1 × 3 + 5 2 × 4 + 5 3 × 5 + 5 4 × 6 + ⋯ = A B \large \dfrac{5}{1 \times 3}+\dfrac{5}{2\times 4}+\dfrac{5}{3 \times 5}+\dfrac{5}{4\times 6}+\cdots = \dfrac{A}{B}

Given that A A and B B are coprime positive integers, find A − B A-B .


The answer is 11.

Anish Harsha by Anish Harsha , 20, India

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2 solutions

Akshat Sharda
Dec 28, 2015

= ∑ n = 1 ∞ 5 n ( n + 2 ) = 5 2 ∑ n = 1 ∞ ( 1 n − 1 n + 2 ) = 5 2 ( ∑ n = 1 ∞ 1 n − ∑ n = 3 ∞ 1 n ) = 5 2 ( 1 + 1 2 ) = 15 4 ⇒ 15 − 4 = 11 \begin{aligned} & = \displaystyle \sum^{\infty}_{n=1}\frac{5}{n(n+2)} \\ & = \frac{5}{2} \displaystyle \sum^{\infty}_{n=1}\left(\frac{1}{n}-\frac{1}{n+2}\right) \\ & = \frac{5}{2} \left(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n}- \displaystyle \sum^{\infty}_{n=3} \frac{1}{n}\right) \\ & = \frac{5}{2}\left(1+\frac{1}{2}\right) \\ & = \frac{15}{4} \\ & \Rightarrow 15-4 = \boxed{11} \end{aligned}

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How does 5/2 come out?

Abu Bakar Khan - 5 years, 5 months ago

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= ∑ n = 1 ∞ 5 n ( n + 2 ) = 5 ∑ n = 1 ∞ 1 n ( n + 2 ) = 5 ∑ n = 1 ∞ 1 2 ( 1 n − 1 n + 2 ) = 5 2 ∑ n = 1 ∞ ( 1 n − 1 n + 2 ) = \displaystyle \sum^{\infty}_{n=1}\frac{5} {n(n+2)} \\ = 5 \displaystyle \sum^{\infty}_{n=1}\frac{1} {n(n+2)} \\ =5 \displaystyle \sum^{\infty}_{n=1}\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right) \\ = \frac{5}{2} \displaystyle \sum^{\infty}_{n=1}\left(\frac{1}{n}-\frac{1}{n+2}\right)

Akshat Sharda - 5 years, 5 months ago

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It makes sense now. Thanks!

Abu Bakar Khan - 5 years, 5 months ago

same way exactly

Kaustubh Miglani - 5 years, 5 months ago

should not the geometric series 1/n diverge as n tends to infinity? thanks

guido barta - 5 years, 5 months ago

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= ( ∑ n = 1 ∞ 1 n ) − ( ∑ n = 3 ∞ 1 n ) = ( 1 + 1 2 + 1 3 + 1 4 + … ) − ( 1 3 + 1 4 + 1 5 + … ) = 1 + 1 2 + ̸ 1 3 + ̸ 1 4 + … − ̸ 1 3 − ̸ 1 4 − ̸ 1 5 − … = 1 + 1 2 = 3 2 =\left(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n}\right)-\left(\displaystyle \sum^{\infty}_{n=3} \frac{1}{n}\right) \\ =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots \right) \\ = 1+\frac{1}{2}+\not{\frac{1}{3}}+\not{\frac{1}{4}}+\ldots- \not{\frac{1}{3}}-\not{\frac{1}{4}}-\not{\frac{1}{5}}-\ldots \\ =1+\frac{1}{2}=\frac{3}{2}

Akshat Sharda - 5 years, 5 months ago

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you are great

guido barta - 5 years, 5 months ago

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@Guido Barta – No problem!

Akshat Sharda - 5 years, 5 months ago
Adams Ayoade
Mar 25, 2016

Woah,solved same way.

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