MathematicsMedium110×since 2002Q3861Let ∑k=110f(a+k)=16(210−1)\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)}k=1∑10f(a+k)=16(210−1) where the function ƒ satisfies ƒ(x + y) = ƒ(x)ƒ(y) for all natural numbers x, y and ƒ(1) = 2. then the natural number 'a' isA2B16C4D3Check answerSkip