Question

The value of $\frac{1^3+2^3+3^3+\_\_\_\_\_\_+{10}^3}{1+2+3+\_\_\_\_\_\_+10}$ ______________.

Answer 1

$\frac{1^3+2^3+3^3+...+{10}^3}{1+2+...+10}$
$\begin{array}{rcl}\mathrm{Since}{1}^3+2^3+...+n^3& =& {\left[\frac{n\left(n+1\right)}{2}\right]}^2\\ & \Rightarrow & 1^3+2^3+...+{10}^3={\left[\frac{10\left(10+1\right)}{2}\right]}^2\\ & =& {\left[\frac{10\times 11}{2}\right]}^2\\ & =& {\left(55\right)}^2\end{array}$
$$\begin{gathered} \mathrm{also}1+2+...+n=\frac{n\left(n+1\right)}{2} \\ \Rightarrow 1+2+...+10=\frac{10\left(11\right)}{2}=\frac{5\times 11}{2}=55 \\ \Rightarrow \frac{1^3+2^3+...+{10}^3}{1+2+..+n}=\frac{{\left(55\right)}^2}{55}=55 \end{gathered}$$