Animated Solution for Mathematics - Binomial Theorem: The positive value of $\lambda$ for which the co-efficient of $x^2$ in the expression $x^2{\left(\sqrt{x}+\frac{\lambda }{x^2}\right)}^{10}$ is 720, is :

Welcome, future engineer! Today, we are going to peel back the layers of a classic binomial theorem problem. We are looking at the expression $x^2{\left(\sqrt{x}+\frac{\lambda }{x^2}\right)}^{10}$.

At first glance, it might look intimidating, but let us break it down together.

The most common mistake students make is diving straight into the binomial expansion of ${\left(\sqrt{x}+\frac{\lambda }{x^2}\right)}^{10}$ and forgetting the $x^2$ sitting outside. Think of that $x^2$ as a gatekeeper.

It is waiting to multiply every single term that comes out of that binomial expansion. If you ignore it, you are essentially solving for the wrong term. We must keep this multiplier in our pocket until the very end.

Now, let us bring out our most powerful tool: the general term formula for $(a+b{)}^n$, which is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$. In our case, $n=10$, $a=x^{1/2}$, and $b=\lambda x^{-2}$.

When we substitute these into the formula, we get the general term for the binomial part:

This is the heart of the expansion.

Now, let us incorporate that external $x^2$. The full term becomes:

Let us clean this up by separating the constants from the variables. The constants are $\left(\genfrac{}{}{0ex}{}{10}{r}\right){\lambda }^r$.

Now, look at the $x$ terms: $x^2\cdot x^{(10-r)/2}\cdot x^{-2r}$. Using the laws of exponents, we add the powers:

This is the net exponent of $x$ in our general term.

The problem asks for the coefficient of $x^2$. This means the net exponent $7-\frac{5r}{2}$ must equal $2$.

Setting them equal, we get:

Subtracting $2$ from both sides gives $5-\frac{5r}{2}=0$, which simplifies to $\frac{5r}{2}=5$. Solving for $r$, we find $r=2$.

Now, we substitute $r=2$ back into our constant part: $\left(\genfrac{}{}{0ex}{}{10}{2}\right){\lambda }^2$. We know this equals $720$.

Calculating $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=45$, we get:

Dividing by $45$, we find ${\lambda }^2=16$. Since the problem asks for the positive value, $\lambda =4$.

You have successfully navigated the trap and solved the problem. Keep this logical flow in mind, and no binomial problem will ever stand in your way!