Question

Let $ƒ : N→R$ be a function such that $ƒ(x + y) = 2ƒ(x) ƒ(y)$ for natural numbers x and y. If $ƒ(1) = 2$, then the value of α for which $\displaystyle\sum_{k=1}^{10} f(α+k)=\frac {512}{3} (2^{20}−1)$ holds, is:

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

4

Answer 2

$ƒ(x + y) = 2ƒ(x)ƒ(y)$ & $ƒ(1) = 2$
$x = y = 1$
$f(x) = 2^{(2x−1)}$
$⇒ f(2)=2^3$
$⇒ f(3)=2^5$
Now,
$\displaystyle\sum_{K=1}^{10}f(α+k) =$ $\frac {512}{3}(2^{20}−1)$

2\displaystyle\sum_{K=1}^{10}f(α)f(k) = $$ \frac {512}{3}(2^{20}−1)

$2f(α)[f(1)+f(2)+⋯+f(10)] =$ $\frac {512}{3}(2^{20}−1)$

$2f(α)[2+2^3+2^5+⋯$upto 10 terms$] =$ $\frac {512}{3}(2^{20}−1)$

$2f(α)⋅2(\frac {2^{20}−1}{4−1}) = \frac {512}{3}(2^{20}−1)$
$ƒ(α) = 128 = 2^{2α} – 1$
$2α – 1 = 7$
$α = 4$

So, the correct option is (C): $4$