Review Note

因式分解公式6
\(a^{3}-b^{3}=\)________.

\(a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\)

\(\begin{align}
I=\lim _{n \rightarrow \infty} \frac{2^3-1}{2^3+1} \times \frac{3^3-1}{3^3+1} \times \frac{4^3-1}{4^3+1} \times \cdots \times \frac{n^3-1}{n^3+1}
\end{align}\)

\(\begin{aligned} & k^3-1=(k-1)\left(k^2+k+1\right) \\ & k^3+1=(k+1)\left(k^2-k+1\right) \\ & \text { 令 } a_k=k-1, b_k=k^2-k+1=k(k-1)+1 \\ & a_{k+2}=k+1, b_{k+1}=(k+1) k+1=k^2+k+1 \\ & \frac{n^3-1}{n^3+1}=\frac{(n-1)\left(n^2+n+1\right)}{(n+1)\left(n^2-n+1\right)}=\frac{a_n b_{n+1}}{a_{n+2} b_n} \\ & I=\lim _{n \rightarrow \infty} \frac{a_2 b_3}{a_4 b_2} \times \frac{a_3 b_4}{a_5 b_3} \times \frac{a_4 b_5}{a_6 b_4} \times \cdots \times \frac{a_n b_{n+1}}{a_{n+2} b_n} \\ & =\lim _{n \rightarrow \infty} \frac{a_2 a_3}{a_{n+1} a_{n+2}} \times \frac{b_{n+1}}{b_2} \\ & =\lim _{n \rightarrow \infty} \frac{1 * 2 *\left(n^2+n+1\right)}{n(n+1)\left(2^2-2+1\right)} \\ & =\frac{2}{3} \end{aligned}\)