It's Harmonic

Algebra Level 4

∑ n = 1 1729 ( a n a n + 1 ) a 1 a 1730 \large \frac{\displaystyle \sum_{n=1}^{1729} \left(a_n a_{n+1} \right)}{a_1 a_{1730}}

If a 1 , a 2 , … , a n a_1,a_2,\ldots,a_n are in a harmonic progression, evaluate the expression above.

This problem is original.


The answer is 1729.

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Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

A Harmonic progression is the series such that 1 a n − 1 a n + 1 = 1 a n − 1 − 1 a n \frac{1}{a_{n}}-\frac{1}{a_{n}+1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}}

a n a n + 1 = a n + 1 − a n 1 a n − 1 a n + 1 a_{n}a_{n+1}=\dfrac{a_{n+1}-a_{n}}{\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+1}}

Then Σ n = 1 1729 a n a n + 1 = 1 1 a n − 1 a n + 1 Σ n = 1 1729 ( a n + 1 − a n ) = 1 1 a n − 1 a n + 1 ( a 1730 − a 1 ) \Sigma_{n=1}^{1729} a_{n}a_{n+1}=\dfrac{1}{\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}} \Sigma_{n=1}^{1729} (a_{n+1}-a_{n})= \dfrac{1}{\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}} (a_{1730}-a_{1})

Finally, we get

1 a 1730 − 1 a 1 1 a 2 − 1 a 1 = 1729 \dfrac{\dfrac{1}{a_{1730}}-\dfrac{1}{a_{1}}}{\dfrac{1}{a_{2}}-\dfrac{1}{a_{1}}}=\boxed{1729}

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