mathop {Lim}limits_{n to infty } frac{{{1^2}n | vedclass

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Question

(A) Let the numerator be $S_n = \sum_{k=1}^{n} k^2(n - k + 1)$.$S_n = (n+1) \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3$.We know that $\sum_{k=1}^{n} k^2

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(A) Let the numerator be $S_n = \sum_{k=1}^{n} k^2(n - k + 1)$.$S_n = (n+1) \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3$.We know that $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.The denominator is $D_n = \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.Thus,the limit is $\lim_{n 	o \infty} \frac{(n+1) \frac{n(n+1)(2n+1)}{6} - \frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)^2}{4}}$.Dividing by $\frac{n^2(n+1)^2}{4}$:$= \lim_{n 	o \infty} \left( \frac{(n+1) n(n+1)(2n+1)}{6} \cdot \frac{4}{n^2(n+1)^2} - 1 ight)$.$= \lim_{n 	o \infty} \left( \frac{4n(n+1)^2(2n+1)}{6n^2(n+1)^2} - 1 ight)$.$= \lim_{n 	o \infty} \left( \frac{4(2n+1)}{6n} - 1 ight) = \frac{8}{6} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

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