The value of 1 × 2 2 + 2 × 3 2 + … + 100 × ( 101 ) 2 1 2 × 2 + 2 2 × 3 + … + 100 2 × 101 is [2024]

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Solution

Q.
The value of $\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots + 100 \times {(101)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots + 100^{2} \times 101}$ is [2024]

1 $\frac{305}{301}$

2 $\frac{306}{305}$

3 $\frac{31}{30}$

4 $\frac{32}{31}$

Ans.
(1)

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

$\Rightarrow \frac{\sum_{n = 1}^{100} n {(n + 1)}^{2}}{\sum_{n = 1}^{100} n^{2} (n + 1)}$

$\Rightarrow \frac{\sum_{n = 1}^{100} (n^{3} + 2 n^{2} + n)}{\sum_{n = 1}^{100} (n^{3} + n^{2})} = \frac{\sum_{n = 1}^{100} n^{3} + \sum_{n = 1}^{100} 2 n^{2} + \sum_{n = 1}^{100} n}{\sum_{n = 1}^{100} n^{3} + \sum_{n = 1}^{100} n^{2}}$

$\Rightarrow \frac{{(\frac{100 (101)}{2})}^{2} + \frac{2 \cdot 100 (101) (201)}{6} + \frac{100 (101)}{2}}{{(\frac{100 (101)}{2})}^{2} + \frac{100 (101) (201)}{6}} = \frac{305}{301}$

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