© All Rights reserved @ LearnWithDash
Step 1: Express the Given Expression in a General Term
We have the expression:
$$x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}.$$
To expand it, we consider the binomial form
$$(a + b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r.$$
Step 2: Identify the General Term in the Expansion
Let
$$a = \sqrt{x} \quad \text{and} \quad b = \frac{\lambda}{x^2},$$
and note that $n = 10.$ The general term (the $(r+1)$-th term) of
$$\left(\sqrt{x} + \frac{\lambda}{x^2}\right)^{10}$$
is
$$T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \left(\frac{\lambda}{x^2}\right)^r.$$
Multiplying by the outside factor $x^2,$ we get:
$$
T_{r+1} = x^2 \binom{10}{r} \left(x^{\frac{10-r}{2}}\right)\left(\lambda^r x^{-2r}\right).
$$
Step 3: Simplify the Power of x
The power of $x$ in the term becomes:
$$
x^2 \cdot x^{\frac{10-r}{2}} \cdot x^{-2r}
= x^{\,2 + \frac{10-r}{2} - 2r}\,.
$$
For this to be $x^2,$ the exponent of $x$ must equal 2:
$$
2 + \frac{10-r}{2} - 2r = 2.
$$
Step 4: Solve for r
Simplify the equation:
$$
\frac{10 - r}{2} - 2r = 0
\quad \Longrightarrow \quad
10 - r = 4r
\quad \Longrightarrow \quad
5r = 10
\quad \Longrightarrow \quad
r = 2.
$$
Step 5: Determine the Required Coefficient
When $r = 2,$ the term contributing to $x^2$ has coefficient:
$$
\binom{10}{2} \lambda^2.
$$
We are given that this coefficient is 720. Hence,
$$
\binom{10}{2} \lambda^2 = 720.
$$
Recall that
$$
\binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45.
$$
So the equation becomes:
$$
45 \lambda^2 = 720
\quad \Longrightarrow \quad
\lambda^2 = 16
\quad \Longrightarrow \quad
\lambda = \pm 4.
$$
Step 6: Conclude the Positive Value of λ
The problem specifies the positive value, so:
$$
\lambda = 4.
$$
Thus, the required positive value of $\lambda$ for which the coefficient of $x^2$ in the given expression is 720 is 4.
