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Let sum(k=1)^(10)f(a+k)=16(2^(10)-1), w...

Let `sum_(k=1)^(10)f(a+k)=16(2^(10)-1),` where the function f satisfies `f(x+y)=f(x)f(y)` for all natural numbers x, y and f(1) = 2. Then, the natural number 'a' is

A

2

B

4

C

3

D

16

Text Solution

Verified by Experts

The correct Answer is:
C
Given, `f(x+y)=f(x)*f(y)`
Let ` f(x)=lambda^(x) " " ["where "lambda gt 0]`
`because f(1)=2 " (given)" `
`therefore lambda =2`
So, `sum_(k=1)^(10)f(a+k)=sum_(k=1)^(10) lambda^(a+k)=lambda^(a)(sum_(k=1)^(10)lambda^(k))`
`=2^(a)[2^(1)+2^(2)+2^(3)+ ... +2^(10)]`
`=2^(a)[(2(2^(10)-1))/(2-1)]`
[by using formula of sum of n-terms of a GP having first term 'a' and common ratio 'r', is
`S_(n)=(a(r^(n)-1))/(r-1)," where " r gt 1]`
` rArr 2^(a+1)(2^(10)-1)=16(2^(10)-1) ("given")`
`rArr 2^(a+1) = 16=2^(4) rArr a+1=4 rArr a=3`
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