SUMO-1

Algebra Level 3

Express 2 3 − 1 2 3 + 1 × 3 3 − 1 3 3 + 1 × 4 3 − 1 4 3 + 1 × ⋯ × 1 6 3 − 1 1 6 3 + 1 \frac{2^3-1}{2^3+1}\times\frac{3^3-1}{3^3+1}\times\frac{4^3-1}{4^3+1}\times\dots\times\frac{16^3-1}{16^3+1} as a fraction in lowest terms.

SUMO: Stanford University Math Organization

1 e \frac{1}{e} 90 131 \frac{90}{131} 80 135 \frac{80}{135} 91 136 \frac{91}{136} 2 3 \frac{2}{3}

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Hana Wehbi
Jun 13, 2018

Note the expression is: ∏ n = 2 k n 3 − 1 n 3 + 1 = ∏ n = 2 k ( n − 1 ) ( n 2 + n + 1 ) ( n + 1 ) ( n 2 − n + 1 ) = ( ∏ n = 2 k n − 1 n + 1 ) ( ∏ n = 2 k n 2 + n + 1 n 2 − n + 1 . ) \prod_{n=2}^{k} \frac{n^3-1}{n^3+1} = \prod_{n=2}^{k}\frac{\Big(n-1\Big)\Big(n^2+n+1\Big)}{\Big(n+1\Big)\Big(n^2-n+1\Big)} =\Bigg (\prod_{n=2}^{k}\frac{n-1}{n+1}\Bigg)\Bigg(\prod_{n=2}^{k}\frac{n^2+n+1}{n^2-n+1}.\Bigg)

Each product telescopes, the final result is: ( 1 × 2 ) k ( k + 1 ) × ( k 2 + k + 1 ) 3 ⟹ for k = 16 the value of the expression is 91 136 . \frac{(1\times2)}{k(k+1)}\times\frac{(k^2+k+1)}{3}\implies \text { for } k=16 \text{ the value of the expression is } \frac{91}{136}.

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