A) 4
B) 3
C) 16
D) 2
Correct Answer: B
Solution :
From the given functional equation : \[\left( x \right)={{2}^{x}}\,\,\,\,\,\,\forall x\in N\] \[{{2}^{a+1}}+{{2}^{a+2}}+....+{{2}^{a+10}}=16({{2}^{10}}-1)\] \[{{2}^{a}}(2+{{2}^{2}}+....+{{2}^{10}})=16({{2}^{10}}-1)\] \[{{2}^{a}}.\frac{2\left( {{2}^{10}}-1 \right)}{1}=16\left( {{2}^{10}}-1 \right)\] \[{{2}^{a+1}}=16={{2}^{4}}\] \[a=3\]You need to login to perform this action.
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