Let \(\mathop \sum \limits_{k = 1}^{10} f\left( {a + k} \right) = 16\left( {{2^{10}} - 1} \right),{\rm{}}\) 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:

Let \(\mathop \sum \limits_{k = 1}^{10} f\left( {a + k} \right) = 16\l

1.	Let \(\mathop \sum \limits_{k = 1}^{10} f\left( {a + k} \right) = 16\left( {{2^{10}} - 1} \right),{\rm{}}\) 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.	16
C.	4
D.	3
Answer» E.

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Let \(\mathop \sum \limits_{k = 1}^{10} f\left( {a + k} \right) = 16\left( {{2^{10}} - 1} \right),{\rm{}}\) 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:

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