Animated Solution for Mathematics - Functions: Let $f:\mathbb{N}\to \mathbb{R}$ be a function such that $f(x+y)=2f(x)f(y)$ for natural numbers $x$ and $y$. If $f(1)=2$, then the value of $\alpha$ for which ${\sum }_{k=1}^{10}f(\alpha +k)=\frac{512}{3}(2^{10}-1)$ holds, is

Welcome, future engineers. Today, we are going to dissect a problem that looks like a daunting wall of abstract algebra but is, in reality, a beautiful staircase leading to a simple, elegant solution. We are dealing with a functional equation: $f(x+y)=2f(x)f(y)$.

Many students freeze when they see this. They think, 'How do I find a function from an equation?' But think of it as a rule of transformation. It tells us how the function behaves when we add inputs. Our mission is to decode this rule and find the general form of $f(n)$.

Decoding the Pattern

We are given the initial condition $f(1)=2$. Let's use our master key, the functional equation, to find the next few terms.

If we set $x=1$ and $y=1$, we get $f(1+1)=2f(1)f(1)$. Substituting our known value, we get $f(2)=2(2)(2)=8$.

Now, let's push further. Set $x=2$ and $y=1$. Then $f(3)=2f(2)f(1)=2(8)(2)=32$.

Look at these results: $f(1)=2^1$, $f(2)=2^3$, $f(3)=2^5$. Do you see the rhythm? The exponents are $1,3,5,\dots$.

This is an arithmetic progression! The general term for these odd numbers is $2n-1$. Thus, we have unlocked the secret: $f(n)=2^{2n-1}$. This is the heartbeat of our problem.

The Summation Challenge

Now that we have $f(n)=2^{2n-1}$, we need to evaluate the summation ${\sum }_{k=1}^{10}f(\alpha +k)$. Let's substitute our general formula into this expression.

We get:

Using the laws of exponents, we can split this into $2^{2\alpha -1}\cdot 2^{2k}$. Notice that $2^{2\alpha -1}$ does not depend on $k$.

It is a constant relative to our summation. We can pull it out of the sigma notation like a magician pulling a rabbit out of a hat:

The Geometric Progression

Now, focus on the summation ${\sum }_{k=1}^{10}{2}^{2k}$. If we expand this, we get $2^2+2^4+2^6+\cdots +2^{20}$.

This is a classic Geometric Progression (G.P.) where the first term $a=2^2=4$ and the common ratio $r=2^2=4$.

The sum of a G.P. is given by $S_n=\frac{a(r^n-1)}{r-1}$. With $n=10$ terms, we have:

The Final Convergence

We are almost there. Let's reassemble our expression: $2^{2\alpha -1}\cdot \frac{4(2^{20}-1)}{3}$. Since $4=2^2$, we combine the powers of two: $2^{2\alpha -1}\cdot 2^2=2^{2\alpha +1}$.

Our expression becomes:

Now, we equate this to the given value. Assuming the corrected RHS for the equation to hold for an integer $\alpha$, we have:

Canceling the common terms, we are left with $2^{2\alpha +1}=512$. Since $512=2^9$, we equate the exponents: $2\alpha +1=9$.

This gives $2\alpha =8$, and finally, $\alpha =4$.

See how the complexity melted away? By breaking the problem into logical phases—decoding the function, identifying the G.P., and simplifying the algebra—we turned a terrifying equation into a simple arithmetic result. Keep this structured approach in your toolkit, and no problem will ever be too big for you.