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

The Correct Option is C

\(ƒ(x + y) = 2ƒ(x)ƒ(y)\) & \(ƒ(1) = 2\)
\(x = y = 1\)
\(f(x) = 2^{(2x−1)}\)
\(⇒ f(2)=2^3 \)
\(⇒ f(3)=2^5\)
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): \(4\)