• # question_answer Let$\sum\limits_{k=1}^{10}{f(a+k)=16}\left( {{2}^{10}}-1 \right),$ where the function ? satisfies $\left( x+y \right)=\left( x \right)\left( y \right)$for all natural numbers x, y and ?(1) = 2. then the natural number 'a' is [JEE Main 9-4-2019 Morning] A) 4                                 B) 3C) 16                               D) 2

From the given functional equation : $\left( x \right)={{2}^{x}}\,\,\,\,\,\,\forall x\in N$ ${{2}^{a+1}}+{{2}^{a+2}}+....+{{2}^{a+10}}=16({{2}^{10}}-1)$ ${{2}^{a}}(2+{{2}^{2}}+....+{{2}^{10}})=16({{2}^{10}}-1)$ ${{2}^{a}}.\frac{2\left( {{2}^{10}}-1 \right)}{1}=16\left( {{2}^{10}}-1 \right)$ ${{2}^{a+1}}=16={{2}^{4}}$ $a=3$