Animated Solution for Mathematics - Binomial Theorem: The maximum value of the term independent of 't' in the expansion of $(t{x}^{1/5}+\frac{(1-x{)}^{1/10}}{t}{)}^{10}$ where $x\in (0,1)$ is:

My dear student, welcome to a journey through one of the most beautiful landscapes in algebra: the Binomial Theorem. Today, we are not just solving a problem; we are uncovering the hidden structure within the expression $(t{x}^{1/5}+\frac{(1-x{)}^{1/10}}{t}{)}^{10}$.

It might look like a daunting mess of variables and fractional powers, but I promise you, there is a rhythm to it. Let us peel back the layers together.

The Hunt for the Independent Term

Every binomial expansion is a collection of terms, and the general term $T_{r+1}$ is our master key. We know that for any expression $(a+b{)}^n$, the general term is given by $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$.

In our case, $n=10$, $a=t{x}^{1/5}$, and $b=\frac{(1-x{)}^{1/10}}{t}$.

When we substitute these into our formula, we get:

Now, here is where the magic happens. We need to isolate the variable $t$. Let us gather all the $t$ terms together.

From the first part, we have $t^{10-r}$. From the second part, because $t$ is in the denominator, it brings a $t^{-r}$ into the mix. When we multiply these, we add the exponents: $10-r+(-r)=10-2r$.

So, our term becomes:

The question asks for the term independent of $t$. This is the moment of truth. For a term to be independent of $t$, the exponent of $t$ must be zero.

Thus, we set $10-2r=0$, which immediately gives us $r=5$. We have found our target!

The Calculus of Optimization

With $r=5$, our term $T_6$ (the sixth term) simplifies beautifully. Substituting $r=5$ into our expression, the $t$ term becomes $t^0=1$.

The $x$ terms become $x^{\frac{10-5}{5}}=x^1$ and $(1-x{)}^{\frac{5}{10}}=(1-x{)}^{1/2}$. So, we are left with:

Now, we enter the realm of calculus. We want to maximize this term. Since $\left(\genfrac{}{}{0ex}{}{10}{5}\right)$ is just a constant, we only need to maximize the function $f(x)=x\sqrt{1-x}$ for $x\in (0,1)$.

Imagine this curve—it starts at zero, rises to a peak, and falls back to zero at $x=1$. To find that peak, we find the derivative $f^{\prime }(x)$ and set it to zero.

Using the product rule, the derivative is:

Setting $f^{\prime }(x)=0$ leads us to:

Multiplying both sides by $2\sqrt{1-x}$ gives $2(1-x)=x$, which simplifies to $2-2x=x$, or $3x=2$. Thus, the maximum occurs at $x=2/3$.

The Final Synthesis

We have done the hard work. Now, we simply substitute $x=2/3$ back into our expression for $T_6$:

Finally, we expand the binomial coefficient $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=\frac{10!}{5!5!}$. Putting it all together, we arrive at our final, elegant result:

Look at that! From a complex binomial expression, we navigated through algebra and calculus to find a precise, beautiful constant. This, my friend, is the power of physics and mathematics. You didn't just solve a problem; you mastered a process. Keep that curiosity alive!