The value of 1 × 2² + 2 × 3² + + 100 × (101)²/1² × 2 + 2² × 3 + + 100² × 101 is [4 Apr 2024 Shift 2]

Correct Answer is : 305301305/301301305

Correct Answer is : 305301305/301301305

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Sequences and Series Part 3

Section: Mathematics

Question : 77 of 109

Marks: +1, -0

The value of

\frac{1 \times 2^2 + 2 \times 3^2 + \dots + 100 \times (101)^2}{1^2 \times 2 + 2^2 \times 3 + \dots + 100^2 \times 101}

Solution: 👈: Video Solution

\frac{1 \times 2^2 + 2 \times 3^2 + \dots + 100 \times (101)^2}{1^2 \times 2 + 2^2 \times 3 + \dots + 100^2 \times 101} = \frac{\sum\limits_{r=1}^{100} r (r+1)^2}{\sum\limits_{r=1}^{100} r^2 (r+1)}

= \frac{\sum\limits_{r=1}^{100} (r^3+2r^2+r)}{\sum\limits_{r=1}^{100} (r^3+r^2)} = \frac{\frac{n(n+1)^2}{2} + \frac{2 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 (2n+1) + 1 \right] }{ \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{2n+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}

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