Baltic Way 1992 · Problem 6

Algebra

Prove that the product of the 99 numbers of the form $\frac{k^{3}-1}{k^{3}+1}$ where $k=2,3, \ldots, 100$ , is greater than $\frac{2}{3}$ .

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Review

Polynomials · Sequences and recurrences

Solution

Solution:

Note that

\frac{k^{3}-1}{k^{3}+1}=\frac{(k-1)\left(k^{2}+k+1\right)}{(k+1)\left(k^{2}-k+1\right)}=\frac{(k-1)\left(k^{2}+k+1\right)}{(k+1)\left((k-1)^{2}+(k-1)+1\right)}

After obvious cancellations we get

\prod_{k=2}^{100} \frac{k^{3}-1}{k^{3}+1}=\frac{1 \cdot 2 \cdot\left(100^{2}+100+1\right)}{100 \cdot 101 \cdot\left(1^{2}+1+1\right)}>\frac{2}{3}

Results from Baltic Way 1992

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