Can we use $a^3+b^3$ or $a^3-b^3$ formula?

Algebra Level 4

$\frac{(2^3-1)(3^3-1)(4^3-1)\cdots(1000^3-1)}{(2^3+1)(3^3+1)(4^3+1)\cdots(1000^3+1)}=\, ?$

Submit your answer to three decimal places.

The answer is 0.666.

1 solution

Chew-Seong Cheong
Jun 23, 2016

Relevant wiki: Telescoping Series - Product

$\begin{aligned} P & = \prod_{n=2}^{1000} \frac {n^3-1}{n^3+1} \\ & = \prod_{n=2}^{1000} \frac {(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \frac {\color{#3D99F6}{\prod_{n=2}^{1000} (n-1)} \prod_{n=2}^{1000} (n^2+n+1)}{\color{#3D99F6}{\prod_{n=2}^{1000} (n+1)} \prod_{n=2}^{1000} (\color{#D61F06}{n^2-n+1})} \quad \quad \small \color{#D61F06}{\text{Note that }(n+1)^2 - (n+1) +1 = n^2 + n +1} \\ & = \frac {\color{#3D99F6}{\prod_{n=1}^{999} n} \prod_{n=2}^{1000} (n^2+n+1)}{\color{#3D99F6}{\prod_{n=3}^{1001} n} \prod_{n=\color{#D61F06}{1}} ^{\color{#D61F06}{999}} (\color{#D61F06}{n^2+n+1})} \\ & = \frac {\color{#3D99F6}{2} \cdot 1001001}{\color{#3D99F6}{1000 \cdot 1001} \cdot \color{#D61F06}{3}} \\ & = \frac {333667} {500500} \\ & \approx \boxed{0.667} \end{aligned}$

Sir how have you changed $(n^2-n+1)$ to $(n^2+n+1)$ ??? Thank you

Chirayu Bhardwaj - 4 years, 11 months ago

Note that $n^{2} - n + 1 = (n - 1)^{2} + (n - 1) + 1$ , so

$\displaystyle\prod_{n=2}^{1000}(n^{2} - n + 1) = \prod_{n=2}^{1000}((n - 1)^{2} + (n - 1) + 1) = \prod_{k=1}^{999}(k^{2} + k + 1)$ ,

where $k = n - 1$ goes from $1$ to $999$ as $n$ goes from $2$ to $1000$ .

Brian Charlesworth - 4 years, 11 months ago

Can we use a 3 + b 3 a^3+b^3 a 3 + b 3 or a 3 − b 3 a^3-b^3 a 3 − b 3 formula?

The answer is 0.666.

1 solution

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Can we use $a^3+b^3$ or $a^3-b^3$ formula?