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

Correct answer: D. (305)/(301) aligned & (1 × 2^2+2 × 3^2+ +100 ×(101)^2)/(1^2 × 2+2^2 × 3+ +100^2 × 101)= ∑_ r=1 ^ 100 r(r+1)^2 ∑_ r=1 ^ 100 r^2(r+1) & = ∑_ r=1 ^ 100 (r^3+2 r^2+r) ∑_ r=1 ^ 100 (r^3+r^2) =(((n(n+1)^2)/(2))+(2 · n(n+1)(2 n+1))/(6)+(n(n+1))/(2))/(((n(n+1))/(2))^2+(n(n+1)(2 n+1))/(6)) & =((n(n+1))/(2)[(n(n+1))/(2)+(2)/(3) ·(2 n+1)+1])/((n(n+1))/(2)[(n(n+1))/(2)+((2 n+1))/(3)]) ; Put n=100 & =((100(101))/(2)+(2)/(3)(201)+1)/((100 × 101)/(2)+(201)/(3))=(5185)/(5117)=(305)/(301) & aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Sequences and Series

The value of (1 × 2^2+2 × 3^2+ +100 ×(101)^2)/(1^2 × 2+2^2 × 3+...

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}

Options

A $\frac{32}{31}$
B $\frac{31}{30}$
C $\frac{306}{305}$
D $\frac{305}{301}$

Correct answer

D. $\frac{305}{301}$

Step-by-step solution

\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}=\frac{\sum_{r=1}^{100} r(r+1)^2}{\sum_{r=1}^{100} r^2(r+1)} \\ & =\frac{\sum_{r=1}^{100}\left(r^3+2 r^2+r\right)}{\sum_{r=1}^{100}\left(r^3+r^2\right)}=\frac{\left(\frac{n(n+1)^2}{2}\right)+\frac{2 \cdot n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}\right)^2+\frac{n(n+1)(2 n+1)}{6}} \\ & =\frac{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3} \cdot(2 n+1)+1\right]}{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{(2 n+1)}{3}\right]} ; \text{ Put n=100 } \\ & =\frac{\frac{100(101)}{2}+\frac{2}{3}(201)+1}{\frac{100 \times 101}{2}+\frac{201}{3}}=\frac{5185}{5117}=\frac{305}{301} \\ & \end{aligned}

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