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 If limnk=0n nCknk(k+3)=eλ then λ=

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detailed solution

Correct option is B

limn→∞ k=0n nCknkk+3=limn→∞ k=0n1k+3⋅nCk⋅1nk                                 =limn→∞ k=0nCk   n⋅1nk∫01xk+2dx                                 =∫01x2limn→∞ k=0nxnkdx                                 =∫01x2⋅limn→∞1+xnndx                                 =∫01x2exdx                                =e−2


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