Determine the term independent of 'x' in the expansion of `(1+x +x^(-2) + x^(-3) )^10`

Answer

To find the term independent of \( x \) in the expansion of \( (1 + x + x^{-2} + x^{-3})^{10} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + x + x^{-2} + x^{-3})^{10} \]
This can be rewritten as:
\[ (1 + x + \frac{1}{x^2} + \frac{1}{x^3})^{10} \]

### Step 2: Combine the terms
We can combine the terms \( x^{-2} \) and \( x^{-3} \) under a common denominator:
\[ = (1 + x + \frac{x + 1}{x^3})^{10} = (1 + x + \frac{1 + x}{x^3})^{10} \]
This simplifies to:
\[ = (1 + x + \frac{1 + x}{x^3})^{10} = (1 + x + \frac{1 + x}{x^3})^{10} \]

### Step 3: Factor out \( x^{-3} \)
We can factor out \( x^{-3} \):
\[ = \frac{1}{x^{30}} (1 + x + x^3)^{10} \]

### Step 4: Expand using the Binomial Theorem
Using the Binomial Theorem, we expand \( (1 + x + x^3)^{10} \):
\[ = \sum_{k=0}^{10} \binom{10}{k} (1)^{10-k} (x + x^3)^k \]
This gives us:
\[ = \sum_{k=0}^{10} \binom{10}{k} (x + x^3)^k \]

### Step 5: Expand \( (x + x^3)^k \)
Now we expand \( (x + x^3)^k \):
\[ = \sum_{j=0}^{k} \binom{k}{j} x^j (x^3)^{k-j} = \sum_{j=0}^{k} \binom{k}{j} x^{j + 3(k-j)} = \sum_{j=0}^{k} \binom{k}{j} x^{3k - 2j} \]

### Step 6: Find the term independent of \( x \)
We need the exponent of \( x \) to be zero:
\[ 3k - 2j = 0 \implies j = \frac{3k}{2} \]
For \( j \) to be an integer, \( k \) must be even. Let \( k = 2m \):
\[ j = 3m \]
Thus, \( 0 \leq 3m \leq 10 \) implies \( m \) can take values \( 0, 1, 2, 3 \).

### Step 7: Calculate the coefficients
We calculate the coefficients for \( m = 0, 1, 2, 3 \):
- For \( m = 0 \): \( k = 0 \), \( j = 0 \) gives \( \binom{10}{0} \binom{0}{0} = 1 \)
- For \( m = 1 \): \( k = 2 \), \( j = 3 \) gives \( \binom{10}{2} \binom{2}{3} = 0 \)
- For \( m = 2 \): \( k = 4 \), \( j = 6 \) gives \( \binom{10}{4} \binom{4}{6} = 0 \)
- For \( m = 3 \): \( k = 6 \), \( j = 9 \) gives \( \binom{10}{6} \binom{6}{9} = 0 \)

### Step 8: Combine the results
Thus, the total coefficient for the term independent of \( x \) is:
\[ 1 + 0 + 0 + 0 = 1 \]

### Step 9: Final term independent of \( x \)
Since we factored out \( x^{-30} \), the term independent of \( x \) in the original expansion is:
\[ \text{Term independent of } x = 1 \]