The coefficient of `x^(-10)` in `(x^2-1/x^3)^10`, is

Options

A. -252
B. 210
C. `-5! `
D. -120

Answer

To find the coefficient of \( x^{-10} \) in the expansion of \( (x^2 - \frac{1}{x^3})^{10} \), we can use the Binomial Theorem. Let's break down the solution step by step.

### Step 1: Identify the general term in the binomial expansion
The Binomial Theorem states that the expansion of \( (a + b)^n \) can be expressed as:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^2 \), \( b = -\frac{1}{x^3} \), and \( n = 10 \). Therefore, the general term \( T_r \) in the expansion is:
\[ T_r = \binom{10}{r} (x^2)^{10-r} \left(-\frac{1}{x^3}\right)^r \]

### Step 2: Simplify the general term
Now, simplify the general term:
\[ T_r = \binom{10}{r} (x^{20 - 2r}) \left(-1\right)^r (x^{-3r}) \]
Combining the powers of \( x \):
\[ T_r = \binom{10}{r} (-1)^r x^{20 - 2r - 3r} = \binom{10}{r} (-1)^r x^{20 - 5r} \]

### Step 3: Set the exponent equal to -10
We need to find the coefficient of \( x^{-10} \). Therefore, we set the exponent equal to -10:
\[ 20 - 5r = -10 \]

### Step 4: Solve for \( r \)
Rearranging the equation:
\[ 20 + 10 = 5r \implies 30 = 5r \implies r = 6 \]

### Step 5: Find the coefficient for \( r = 6 \)
Now we substitute \( r = 6 \) back into the general term to find the coefficient:
\[ T_6 = \binom{10}{6} (-1)^6 x^{20 - 5 \cdot 6} \]
Since \( (-1)^6 = 1 \):
\[ T_6 = \binom{10}{6} x^{-10} \]
The coefficient we need is \( \binom{10}{6} \).

### Step 6: Calculate \( \binom{10}{6} \)
Using the formula for combinations:
\[ \binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} \]
Calculating this:
\[ \binom{10}{6} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210 \]

### Final Answer
Thus, the coefficient of \( x^{-10} \) in the expansion of \( (x^2 - \frac{1}{x^3})^{10} \) is **210**.

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