Jose's cubic

Algebra Level 4

Find the positive integer value of n n such that

n 3 − 3 n 3 + n 3 − 4 n 3 + n 3 − 5 n 3 + … + 4 n 3 + 3 n 3 = 169. \frac{n^3 - 3}{n^3} + \frac{n^3 - 4}{n^3} + \frac{n^3 - 5}{n^3} + \ldots + \frac{4}{n^3} + \frac{3}{n^3} = 169.

This problem is posed by Jose N.

Details and assumptions

The pattern in the above sequence is that the numerator decreases by 1, while the denominator is constant.


The answer is 7.

Arron Kau by Arron Kau

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1 solution

Pedro Ramirez
Jun 26, 2014

Rewrite the LHS as

− 3 + 1 + 2 + 3 + 4 + . . . + ( n 2 − 5 ) + ( n 2 − 4 ) + ( n 2 − 3 ) n 3 \displaystyle \frac {-3 + 1 + 2 + 3 + 4 + ... + (n^{2} - 5) + (n^{2} - 4) + (n^{2} - 3)}{n^{3} } .

Then we add terms and obtain

− 3 + ( n 2 − 3 ) ( n 2 − 2 ) 2 n 3 = ( n 2 − 3 ) ( n 2 − 2 ) − 6 2 n 3 = n 6 − 5 n 3 2 n 3 = n 3 − 5 2 \displaystyle \frac{-3 + \frac{(n^{2} - 3)(n^{2} - 2)}{2 }}{n^3} = \frac { (n^{2} - 3)(n^{2} - 2) - 6 } { 2n^{3} } = \frac {n^{6} - 5n^{3}} {2n^{3}} = \frac {n^{3} - 5} {2} ,

since n ≠ 0 n\neq 0 .

Then n 3 − 5 2 = 169 ⇒ n 3 = 343 \displaystyle \frac {n^{3} - 5} {2} = 169 \Rightarrow n^{3} = 343 . Therefore n = 7 n = 7 .

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