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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

16

B

3

C

4

D

2

Text Solution

Verified by Experts

The correct Answer is:
B
`f(x+y)=f(x)*f(y) rArr f(x)=a^(x)`
`f(1)=2 rArr f(x)=2^(x)`
`underset(k=1)overset(10)sum f(a+k)=2^(a)[2+2^(2)+...+2^(10)]=(2^(a+1)(z^(10)-1))/(2-1)`
Also, `2^(a+1)(2^(10)-1)=16(2^(10)-1)`
`rArr a+1=4 rArr a=3`
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