Compute \[\frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \frac{1}{4^3 - 4} + \dots + \frac{1}{10^3 - 10}.\]
+9519 You can just use a calculator, but here's a calculation trick:
The denominator looks like \(n^3 - n = n(n - 1)(n + 1)\) we can perform partial fractions on \(\dfrac1{n(n-1)(n+1)}\) to get \(\dfrac12 \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12 \left(\dfrac1n - \dfrac1{n + 1}\right)\). The original expression is \(\displaystyle \sum_{n=2}^{10}\left(\dfrac12 \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12\left(\dfrac1n - \dfrac1{n + 1}\right)\right)=\dfrac12\displaystyle \sum_{n=2}^{10} \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12\displaystyle \sum_{n=2}^{10} \left(\dfrac1n - \dfrac1{n + 1}\right)\). Now note that each sum is a telescoping sum, so the answer is \(\dfrac12 \left(1 - \dfrac1{10}\right) - \dfrac12 \left(\dfrac12 - \dfrac1{11}\right)\). You can simplify it by hand or by calculator.
