JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

\alpha=\lim \limits_{{n \rightarrow \infty}}\left(\frac{1^2+2^2+\cdots+n^2}{n^3}\right)

and

\beta=\lim \limits_{{n \rightarrow \infty}}\left[\frac{\left(1^3-1^2\right)+\left(2^3-2^2\right)+\left(3^3-3^2\right)+\cdots+\left(n^3-n^2\right)}{n^4}\right]

Then

View answer and explanation

Correct option: D

3 \alpha=4 \beta

Option D (

3 \alpha = 4 \beta

) is correct because after evaluating the limits, the relationship between

\alpha

and

\beta

satisfies this equation.

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