JEE Mains · Maths · STD 11 - 8. sequence and series
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
- A \(\frac{306}{305}\)
- B \(\frac{305}{301}\)
- C \(\frac{32}{31}\)
- D \(\frac{31}{30}\)
Answer & Solution
Correct Answer
(B) \(\frac{305}{301}\)
Step-by-step Solution
Detailed explanation
\( \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101}=\frac{\sum_{\mathrm{r}=1}^{100} \mathrm{r}(\mathrm{r}+1)^2}{\sum_{\mathrm{r}=1}^{100} \mathrm{r}^2(\mathrm{r}+1)} \)…
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