Question

Let \(\mathop \sum \limits_{k = 1}^{10} f\left( {a + k} \right) = 16\left( {{2^{10}} - 1} \right),{\rm{}}\) 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

Answer 1

Correct Answer - Option 4 : 3

Given f(1) = 2 and f(x + y) = f(x)·f(y)

at x = 1,y =1⇒ f(2) = f(1)·f(1) = 2²

x = 2, y =1 ⇒ f(3) = f(2)·f(1) = 2³

…… f(n) = 2ⁿ

Now \(\mathop \sum \limits_{{\rm{k}} = 1}^{10} {\rm{f}}\left( {{\rm{a}} + {\rm{k}}} \right) = 16\left( {{2^{10}} - 1} \right)\) 

⇒ f(a+1) + f(a+2) + … + f(a+10) =16(2¹⁰- 1)

⇒ 2^a+1 + 2^a+2 + … + 2^a+10 = 16(2¹⁰ – 1)

⇒ 2^a [2¹ + 2² + … + 2¹⁰] = 16(2¹⁰ – 1)

\(\Rightarrow {2^a}\left[ {\frac{{2\left( {{2^{10}} - 1} \right)}}{{2 - 1}}} \right] = 16\left( {{2^{10}} - 1} \right)\)

⇒ 2^a+1 = 16

⇒ 2^a+1 = 2⁴

⇒ a + 1 = 4

⇒ a = 3

Therefore, the natural number 'a' is 3.