Find the natural number ' a ' for which k = 1 n f ( a + k ) = 16 ( 2 n - 1 ) , where f ( x ) = 2 x .

Question

Find the natural number $' a'$ for which $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1),$ where $f (x) = 2^{x}$ .

Smart Achievers · Accepted Answer

(3)

$f (x) = 2^{x} for all x \in N$

$∴$ $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1) \Rightarrow \sum_{k = 1}^{n} 2^{a + k} = 16 (2^{n} - 1)$

$\Rightarrow \sum_{k = 1}^{n} 2^{a} \cdot 2^{k} = 16 (2^{n} - 1) \Rightarrow 2^{a} (\sum_{k = 1}^{n} 2^{k}) = 16 (2^{n} - 1)$

$\Rightarrow 2^{a} (2 + 2^{2} + \dots + 2^{n}) = 16 (2^{n} - 1)$

$\Rightarrow 2^{a} {2 [\frac{2^{n} - 1}{2 - 1}]} = 16 (2^{n} - 1)$

$\Rightarrow 2^{a + 1} (2^{n} - 1) = 16 (2^{n} - 1)$

$\Rightarrow 2^{a + 1} = 2^{4} \Rightarrow a + 1 = 4 \Rightarrow a = 3$

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