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Algebra Level 4

1 1 × 7 + 1 4 × 10 + 1 7 × 13 + 1 10 × 16 + 1 13 × 19 + 1 16 × 22 + ⋯ = A B \dfrac{1}{1 \times 7}+\dfrac{1}{4 \times 10}+\dfrac{1}{7 \times 13}+\dfrac{1}{10 \times 16}+\dfrac{1}{13 \times 19}+\dfrac{1}{16 \times 22}+\cdots =\dfrac{A}{B}

Find A + B A+B where A A and B B are coprime positive integers.


The answer is 29.

Akshat Sharda by Akshat Sharda , 21, India

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2 solutions

Akshat Sharda
Nov 24, 2015

1 7 + 1 40 + 1 91 + 1 160 + … = 1 1 × 7 + 1 4 × 10 + 1 7 × 13 + 1 10 × 16 + … \displaystyle \frac{1}{7}+\frac{1}{40}+\frac{1}{91}+\frac{1}{160}+\ldots = \frac{1}{1\times 7}+\frac{1}{4\times 10}+\frac{1}{7\times 13}+\frac{1}{10\times 16}+\ldots

We know 1 n ( n + 6 ) = 1 6 ( 1 n − 1 n + 6 ) \displaystyle \frac{1}{n(n+6)}=\frac{1}{6}\left(\frac{1}{n}-\frac{1}{n+6}\right) by partial fractions.

1 6 ( 1 − 1 7 ) + 1 6 ( 1 4 − 1 10 ) + 1 6 ( 1 7 − 1 13 ) + … \displaystyle \frac{1}{6}\left(1-\frac{1}{7}\right)+\frac{1}{6}\left(\frac{1}{4}-\frac{1}{10}\right)+\frac{1}{6}\left(\frac{1}{7}-\frac{1}{13}\right)+\ldots

1 6 ( 1 − 1 7 + 1 4 − 1 10 + 1 7 − 1 13 + … ) \displaystyle \frac{1}{6}\left(1-\cancel{\frac{1}{7}}+\frac{1}{4}-\cancel{\frac{1}{10}}+\cancel{\frac{1}{7}}-\cancel{\frac{1}{13}}+\ldots \right)

1 6 ( 1 + 1 4 ) = 5 24 \displaystyle \frac{1}{6}\left(1+\frac{1}{4}\right)=\frac{5}{24}

A + B = 29 A+B=\boxed{29}

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Moderator note:

Slight typo in the third line ( = = instead of + + ), but otherwise this is good.

Generally, people do not say " + … ∞ + \ldots \infty ", but just " + … + \ldots " to indicate a sum to infinity.

In the question, can you clarify what the sequence of terms are?

Calvin Lin Staff - 5 years, 6 months ago

I've lightly edited and beautified your solution so other solvers can more easily absorb it :) As the Challenge Master has said, it is best not to attach ∞ \infty to the end of an indefinite sum.

Jake Lai - 5 years, 6 months ago
Nowras Otmen
Dec 3, 2015

I had a similar solution, just used a different sequence:

∑ i = 0 ∞ 1 ( 3 i + 1 ) ( 3 i + 7 ) = ∑ i = 0 ∞ 1 6 ( 1 3 i + 1 − 1 3 i + 7 ) \displaystyle\sum_{i=0}^\infty \dfrac {1}{(3i+1)(3i+7)}=\displaystyle\sum_{i=0}^\infty \dfrac{1}{6}\left( \dfrac{1}{3i+1}-\dfrac{1}{3i+7}\right) by partial fractions; ∑ i = 0 ∞ 1 6 ( 1 3 i + 1 − 1 3 i + 7 ) = 1 6 ( 1 − 1 7 + 1 4 − 1 10 + 1 7 − 1 13 + 1 10 − 1 16 + 1 13 − 1 19 + . . . ) \displaystyle\sum_{i=0}^\infty \dfrac{1}{6}\left( \dfrac{1}{3i+1}-\dfrac{1}{3i+7}\right)=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{10}-\dfrac{1}{16}+\dfrac{1}{13}-\dfrac{1}{19}+...\right)

After cancelling out the terms, we are left with:

∑ i = 0 ∞ 1 6 ( 1 3 i + 1 − 1 3 i + 7 ) = 1 6 ( 1 + 1 4 ) \displaystyle\sum_{i=0}^\infty \dfrac{1}{6}\left( \dfrac{1}{3i+1}-\dfrac{1}{3i+7}\right)=\dfrac{1}{6}\left(1+\dfrac{1}{4}\right)

∑ i = 0 ∞ 1 6 ( 1 3 i + 1 − 1 3 i + 7 ) = 5 24 \displaystyle\sum_{i=0}^\infty \dfrac{1}{6}\left( \dfrac{1}{3i+1}-\dfrac{1}{3i+7}\right)=\dfrac{5}{24}

5 24 = A B \dfrac{5}{24}=\dfrac{A}{B}

∴ A + B = 5 + 24 = 29 \therefore A+B=5+24=29

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