Question

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Answer 1

Answer:

  B.  50

Step-by-step explanation:

  $\displaystyle S_n=\sum_{k=1}^n{\left(k^2\dfrac{42}{n^3}+k\dfrac{12}{n^2}+\dfrac{30}{n}\right)}\\\\=\dfrac{42}{n^3}\sum_{k=1}^n{k^2}+\dfrac{12}{n^2}\sum_{k=1}^n{k}+\dfrac{30}{n}\sum_{k=1}^n{1}\\\\=\dfrac{42n(n+1)(2n+1)}{6n^3}+\dfrac{12n(n+1)}{2n^2}+\dfrac{30n}{n}\\\\=14+\dfrac{21}{n}+\dfrac{7}{n^2}+6+\dfrac{6}{n}+30=50+\dfrac{27n+7}{n^2}$

As n gets large, the fraction disappears, so the limit is 50.

Answer 2

Answer:

50

Step-by-step explanation:

42/n³∑k²+12/n²∑k+30/n∑1

=42/n³[n(n+1)(2n+1)/6]+12/n²[n(n+1)/2]+30/n [n]

=7n(n+1)(2n+1)/n³+6n(n+1)/n²+30

=7(n+1)(2n+1)/n²+6(n+1)/n+30

=[7(2n²+3n+1)+6(n²+n)+30n²]/n²

=[14n²+21n+7+6n²+6n+30n²]/n²

=[50n²+27n+7]/n²

=[50+27/n+7/n²]

→50 as n→∞

because 1/n,1/n²→0 as n→∞