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
Solution
<p>$$\begin{aligned}
& \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n(n+1)^2}{\sum_\limits{n=1}^{100} n^2(n+1)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{\sum_\limits{n=1}^{100} n^3+2 n^2+n}{\sum_\limits{n=1}^{100} n^3+n^2} \\
& =\frac{\left(\frac{100(101)}{2}\right)^2+\frac{2 \cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{\left(\frac{100(101)}{2}\right)^2+\frac{100(101)(201)}{6}} \\
& =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
This question is part of PrepWiser's free JEE Main question bank. 209 more solved questions on Sequences and Series are available — start with the harder ones if your accuracy is >70%.