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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:
(1) 2
(2) 16
(3) 4
(4) 3
Type: MCQ
$f(x) = 2^x$, $2^a \cdot 2(2^{10}-1) = 16(2^{10}-1)$, $a = 3$
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