f(a+1) + f(a+2) + f(a+3) + â€¦..+f(a+n) = 16 (2^n -1)————–1)

putting n =1

f(a+1) = 16

f(a)*f(1) = 16

or f(a) = 8 as f(1) =2

Putting n =2 in eq 1)

F(a+1) + f(a+2) =16*(2^2 -1)

16 + f(a)*f(2) = 16*3 = 48

8*(f2) = 48 -16 = 32

Or f(2) = 4

By putting n = 3 , in eq 1 we get,

f(a+1) + f(a+2) + f(a+3) = 16*(2^3 -1)

16 + 32 + f(a)*f(3) = 16*7 = 112

F(a)*f(3) = 112 â€“ 16 â€“ 32 = 64

8*f(3) = 64

F(3) = 8

As we know f(a) = 8 so a = 3

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