Question:
Let
\(π_π =β_{k=1}^n\frac{1+k2^k}{4^{k-1}},n=1,2,...\)
Then, limπββ ππ equals (round off to two decimal places)
Let
\(π_π =β_{k=1}^n\frac{1+k2^k}{4^{k-1}},n=1,2,...\)
Then, limπββ ππ equals (round off to two decimal places)
\(π_π =β_{k=1}^n\frac{1+k2^k}{4^{k-1}},n=1,2,...\)
Then, limπββ ππ equals (round off to two decimal places)
Updated On: Oct 1, 2024
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Correct Answer: 9.32
Solution and Explanation
The correct answer is: 9.32 to 9.34 (approx)
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