Question

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Answer 1

50

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→∞

Answer 2

B. 50

Explanation:

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

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