mathop {lim }limits_{n to infty } left[ {frac | vedclass

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Question

(C) The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2

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$\frac{1}{4}$

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(C) The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} ight]^2 = \frac{n^2(n+1)^2}{4}$.Substituting this into the limit expression:$\mathop {\lim }\limits_{n 	o \infty } \frac{n^2(n+1)^2}{4n^4} = \mathop {\lim }\limits_{n 	o \infty } \frac{n^2(n^2 + 2n + 1)}{4n^4}$.$= \mathop {\lim }\limits_{n 	o \infty } \frac{n^4 + 2n^3 + n^2}{4n^4}$.Divide numerator and denominator by $n^4$:$= \mathop {\lim }\limits_{n 	o \infty } \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4} = \frac{1 + 0 + 0}{4} = \frac{1}{4}$.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}{{{n^4}}}} \right] = $

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