Key Concept: The Binomial Theorem and General Term

The foundation for solving this problem lies in the Binomial Theorem, which provides a systematic way to expand expressions of the form $(a+b)^n$. The general term, often denoted as $T_{r+1}$, in the expansion of $(a+b)^n$ is given by:

$T_{r+1} = {}^n C_r a^{n-r} b^r$

where:

• $n$ is the power to which the binomial is raised.
• $r$ is the index of the term (starting from $r=0$ for the first term, $r=1$ for the second, and so on).
• ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, representing the number of ways to choose $r$ items from a set of $n$ items.
• $a$ is the first term of the binomial.
• $b$ is the second term of the binomial.

Problem Breakdown and Setting Up the General Term

We are given the expression $x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$ and asked to find the positive value of $\lambda$ such that the coefficient of $x^2$ is 720.

The expression has two main parts:

1. An external multiplier: $x^2$
2. A binomial expansion: $\left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$

Our strategy is to first find the general term of the binomial expansion, then multiply it by the external $x^2$ term, and finally identify the power of $x$ to determine the coefficient.

For the binomial part $\left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$:

• The power $n = 10$.
• The first term $a = \sqrt{x} = x^{1/2}$. (We express $\sqrt{x}$ as $x^{1/2}$ for easier algebraic manipulation of exponents).
• The second term $b = \frac{\lambda}{x^2} = \lambda x^{-2}$. (Similarly, $1/x^2$ is expressed as $x^{-2}$).

Step 1: Derive the General Term of the Binomial Expansion

Substitute $n=10$, $a=x^{1/2}$, and $b=\lambda x^{-2}$ into the general term formula $T_{r+1} = {}^n C_r a^{n-r} b^r$:

$T_{r+1} = {}^{10} C_r \left( x^{1/2} \right)^{10-r} \left( \lambda x^{-2} \right)^r$

Now, simplify the terms involving $x$ and $\lambda$:

• For the first part: $\left( x^{1/2} \right)^{10-r} = x^{\frac{1}{2}(10-r)} = x^{\frac{10-r}{2}}$
• For the second part: $\left( \lambda x^{-2} \right)^r = \lambda^r \cdot (x^{-2})^r = \lambda^r x^{-2r}$

Substitute these back into the general term: $T_{r+1} = {}^{10} C_r \cdot x^{\frac{10-r}{2}} \cdot \lambda^r \cdot x^{-2r}$

Combine the powers of $x$ using the rule $x^A \cdot x^B = x^{A+B}$: $T_{r+1} = {}^{10} C_r \lambda^r x^{\frac{10-r}{2} - 2r}$ This expression represents the general term within the binomial expansion $\left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$.

Step 2: Incorporate the External Term to Find the Overall General Term

The full expression is $x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$. We must multiply the general term found in Step 1 by the external $x^2$ term:

$T_{r+1}^{\text{overall}} = x^2 \cdot \left( {}^{10} C_r \lambda^r x^{\frac{10-r}{2} - 2r} \right)$

Again, combine the powers of $x$: $T_{r+1}^{\text{overall}} = {}^{10} C_r \lambda^r x^{2 + \frac{10-r}{2} - 2r}$ This is the general term for the entire given expression.

Step 3: Determine the Value of 'r' for the Coefficient of $x^2$

We are looking for the coefficient of $x^2$, which means the power of $x$ in the overall general term must be equal to 2. So, we set the exponent of $x$ equal to 2:

$2 + \frac{10-r}{2} - 2r = 2$

To solve for $r$, first, subtract 2 from both sides of the equation: $\frac{10-r}{2} - 2r = 0$

Next, eliminate the fraction by multiplying the entire equation by 2: $2 \cdot \left( \frac{10-r}{2} \right) - 2 \cdot (2r) = 2 \cdot 0$ $10-r - 4r = 0$

Combine the terms involving $r$: $10 - 5r = 0$

Isolate $r$ by moving $5r$ to the right side of the equation: $10 = 5r$

Finally, divide by 5 to find the value of $r$: $r = \frac{10}{5} = 2$ This tells us that the term containing $x^2$ corresponds to $r=2$.

Step 4: Calculate the Coefficient of $x^2$ in terms of $\lambda$

Now that we have $r=2$, we can substitute this value back into the coefficient part of our overall general term, which is ${}^{10} C_r \lambda^r$.

Substituting $r=2$: $\text{Coefficient of } x^2 = {}^{10} C_2 \lambda^2$

Let's calculate the binomial coefficient ${}^{10} C_2$: ${}^{10} C_2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9 \times 8!}{ (2 \times 1) \times 8!}$ Cancel out $8!$ from the numerator and denominator: ${}^{10} C_2 = \frac{10 \times 9}{2} = 5 \times 9 = 45$

So, the coefficient of $x^2$ in the given expression is $45 \lambda^2$.

Step 5: Solve for $\lambda$

The problem states that the coefficient of $x^2$ is 720. We can now set up an equation:

$45 \lambda^2 = 720$

To find $\lambda^2$, divide both sides by 45: $\lambda^2 = \frac{720}{45}$ Performing the division: $\lambda^2 = 16$

Take the square root of both sides to find $\lambda$: $\lambda = \pm \sqrt{16}$ $\lambda = \pm 4$

The problem explicitly asks for the positive value of $\lambda$. Therefore, $\lambda = 4$.

Common Mistakes and Educational Tips:

• Ignoring the external $x^2$ term: A frequent mistake is to only apply the binomial theorem to $\left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$ and forget to multiply the resulting terms by the $x^2$ outside. Always account for all $x$ terms in the entire expression when finding the overall exponent.
• Exponent rules: Be meticulous with manipulating exponents, especially when dealing with fractions ($\sqrt{x} = x^{1/2}$) and negative powers ($\frac{1}{x^2} = x^{-2}$). A small error here can lead to an incorrect value of $r$. Remember $x^a \cdot x^b = x^{a+b}$ and $(x^a)^b = x^{ab}$.
• Calculation of Binomial Coefficients: Ensure accurate calculation of ${}^n C_r$. Write out the factorial expansion to avoid mistakes.
• Reading the question carefully: Pay close attention to specific conditions, such as "positive value of $\lambda$". If not specified, both $\pm 4$ would be valid mathematical solutions.

Summary and Key Takeaway

This problem is a classic application of the Binomial Theorem that requires careful algebraic manipulation of exponents. The key is to correctly set up the general term for the entire expression, combine all powers of $x$, equate the resulting exponent to the desired power (in this case, 2 for $x^2$), and then solve for the unknown variable. Understanding how each part of the expression contributes to the final power of $x$ is crucial.