Middle term in the expansion of `(x^(2)-2x)^(10)` will be -

Options

A. `.^(10)C_(4)x^(17)2^(4)`
B. `-.^(10)C_(5)2^(5)x^(15)`
C. `-.^(10)C_(4)2^(4)x^(17)`
D. `.^(10)C_(5)2^(4)x^(15)`

Answer

To find the middle term in the expansion of \((x^2 - 2x)^{10}\), we can follow these steps:

### Step 1: Identify the number of terms in the expansion
The number of terms in the expansion of \((x^2 - 2x)^{10}\) can be determined using the binomial theorem. The total number of terms in the expansion is given by \(n + 1\), where \(n\) is the exponent. Here, \(n = 10\), so the number of terms is:
\[ 10 + 1 = 11 \]

### Step 2: Determine the middle term
Since there are 11 terms (which is odd), the middle term is the \(6^{th}\) term. The \(k^{th}\) term in the expansion of \((a + b)^n\) is given by:
\[ T_{k} = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
In our case, \(a = x^2\), \(b = -2x\), and \(n = 10\). For the \(6^{th}\) term, \(k = 6\).

### Step 3: Calculate the \(6^{th}\) term
Using the formula for the \(k^{th}\) term:
\[ T_{6} = \binom{10}{6-1} (x^2)^{10-(6-1)} (-2x)^{6-1} \]
This simplifies to:
\[ T_{6} = \binom{10}{5} (x^2)^{5} (-2x)^{5} \]

### Step 4: Simplify the expression
Now we calculate each part:
1. \(\binom{10}{5} = 252\)
2. \((x^2)^{5} = x^{10}\)
3. \((-2x)^{5} = -32x^{5}\)

Putting it all together:
\[ T_{6} = 252 \cdot x^{10} \cdot (-32x^{5}) = 252 \cdot (-32) \cdot x^{15} \]
Calculating \(252 \cdot (-32)\):
\[ 252 \cdot (-32) = -8064 \]
Thus, the \(6^{th}\) term is:
\[ T_{6} = -8064 x^{15} \]

### Final Answer
The middle term in the expansion of \((x^2 - 2x)^{10}\) is:
\[ -8064 x^{15} \]

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