The value of [(1 2^2+2 3^2++100 (101)^2)/(1^2 2+2^2 3+ .+100^2 101)] is

Question

The value of $$\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$$ is

Accepted Answer

$\begin{gathered} \frac{1	imes {2}^{2}+2	imes {3}^{2}+\ldots +100	imes (101{)}^{2}}{{1}^{2}	imes 2+{2}^{2}	imes 3+\ldots +{100}^{2}	imes 101} \ \Rightarrow \frac{\sum _{n=1}^{100}n(n+1{)}^{2}}{\sum _{n=1}^{100}{n}^{2}(n+1)} \end{gathered}$

$\begin{gathered} \Rightarrow \frac{\sum _{n=1}^{100}{n}^{3}+2{n}^{2}+n}{\sum _{n=1}^{100}{n}^{3}+{n}^{2}} \ =\frac{{(\frac{100(101)}{2})}^{2}+\frac{2\cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{{(\frac{100(101)}{2})}^{2}+\frac{100(101)(201)}{6}} \ =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301} \end{gathered}$

Solutions · 1