The value of lim_n 0.05em ( _k=1^n 0.05em k^3+6k^2+11k+5 (k+3 ) ! ) is:

Question

Accepted Answer

Correct answer: D. (d) limn 0.05em 0.05em 0.05em k^3+6k^2+11k+5 (k+3 ) ! = limn 0.05em 0.05em k^3+6k^2+11k+6-1 (k+3 ) ! = limn 0.05em 0.05em (k+1 ) (k+2 ) (k+3 )-1 (k+3 ) ! = limn 0.05em 0.05em 0.05em (k+1 ) (k+2 ) (k+3 ) (k+3 ) !- 1 (k+3 ) ! = limk=1 0.05em 0.05em 0.05em( 1k !- 1 (k+3 ) !) = lim_k=1 0.05em( 11 !+ 12 !+ 13 !+ 14 ! ⋯+ 1n !- 14 !- 15 !- 16 ! ⋯- 1 (n+3 ) !) = 11+ 12+ 16= 106= 53

$The value of lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k ^{3} + 6 k ^{2} + 11 k + 5}{( k + 3 ) !}) is:$

Solution

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