InterviewSolution
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InterviewSolution
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Let f be a real valued function satisfying `f(x+y)=f(x)f(y) ` for all `x, y in R ` such that f(1)=2 . If ` sum_(k=1)^(n)f(a+k)=16(2^(n)-1) `, then a=A. 3B. 4C. 2D. none of these |
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Answer» Correct Answer - A From illustration 11 , we have `f(x)=[f(1)]^(x)=2^(x) ` for all ` x in R ` `: . sum_(k=1)^(n)f(a+k)=16(2^(n)-1)` `implies sum_(k=1)^(n)2^(a+k)=16(2^(n)-1)` `implies 2^(a) sum_(k=1)^(n)2^(k)=16(2^(n)-1)` `implies 2^(a)xx2 ((2^(n)-1)/(2-1))=16(2^(n)-1)implies 2^(a+1)=2^(4)implies a=3` |
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