Let Σ¹⁰ₖ=₁f(a+k) = 16(2¹⁰-1), 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?
4
16
3
2
Solution:
Correct option is B.3From the given functional equation:f(x)=2x∀x∈N2a+1+2a+2+2a+10=16(210−2;a(2+22+...+210)=16(2102a⋅2⋅(210−1;)1=16(210−1;)2a+1=16=24a=3.