The value of lim(n→∞) [ Σ(k=1 to n) (2k-1) + 8n ] / [ Σ(k=1 to n) (2k-1) + 4n ] is equal to:
The value of lim(n→∞) [ Σ(k=1 to n) (2k-1) + 8n ] / [ Σ(k=1 to n) (2k-1) + 4n ] is equal to:
Option 1 -
2 - logₑ(2/3)
Option 2 -
1 + 2logₑ(3/2)
Option 3 -
5 + logₑ(2/3)
Option 4 -
3 + 2logₑ(2/3)
Asked by Shiksha User
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1 Answer
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Correct Option - 2
Detailed Solution:lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
= (1 + 2ln6) - (0 + 2ln4) = 1 + 2ln (6/4) = 1 + 2ln (3/2).
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