Coefficient of 'x' in the expansion of `(x^m+1/x^2)^(22)` is 1540 and m is a natural number. Find m=?

Answer

To solve the problem of finding the natural number \( m \) such that the coefficient of \( x \) in the expansion of \( (x^m + \frac{1}{x^2})^{22} \) is 1540, we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^m \), \( b = \frac{1}{x^2} \), and \( n = 22 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{22}{r} (x^m)^{22-r} \left(\frac{1}{x^2}\right)^r = \binom{22}{r} x^{m(22-r)} x^{-2r} \]
This simplifies to:
\[ T_{r+1} = \binom{22}{r} x^{22m - mr - 2r} \]

### Step 2: Set the Power of \( x \) Equal to 1
To find the coefficient of \( x \), we need the exponent of \( x \) to be 1:
\[ 22m - mr - 2r = 1 \]

### Step 3: Rearrange the Equation
Rearranging the equation gives:
\[ 22m - mr - 2r = 1 \implies 22m - (m + 2)r = 1 \]

### Step 4: Find the Coefficient
We know that the coefficient \( \binom{22}{r} \) must equal 1540. So we have:
\[ \binom{22}{r} = 1540 \]

### Step 5: Calculate \( \binom{22}{r} \)
We can calculate \( \binom{22}{r} \) for various values of \( r \) to find which one gives 1540.

1. For \( r = 3 \):
\[ \binom{22}{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = \frac{9240}{6} = 1540 \]
This works, so we can use \( r = 3 \).

### Step 6: Substitute \( r \) Back into the Equation
Substituting \( r = 3 \) into the rearranged equation:
\[ 22m - (m + 2) \cdot 3 = 1 \implies 22m - 3m - 6 = 1 \implies 19m - 6 = 1 \]
\[ 19m = 7 \implies m = \frac{7}{19} \]
This is not a natural number.

### Step 7: Try \( r = 19 \)
Now, let's try \( r = 19 \):
\[ \binom{22}{19} = \binom{22}{3} = 1540 \]
Substituting \( r = 19 \) into the equation:
\[ 22m - (m + 2) \cdot 19 = 1 \implies 22m - 19m - 38 = 1 \implies 3m - 38 = 1 \]
\[ 3m = 39 \implies m = 13 \]

### Conclusion
The value of \( m \) is \( 13 \).