Key Concept: The General Term in Binomial Expansion

The core concept for solving this problem is the Binomial Theorem, specifically the formula for the general term (or the $(r+1)^{th}$ term) in the expansion of $(a+b)^n$. This formula is: $T_{r+1} = {^nC_r} a^{n-r} b^r$ where ${^nC_r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, representing the number of ways to choose $r$ items from a set of $n$ items.

Step 1: Identify the components of the given binomial expression

We are given the expression ${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$. By comparing this to the standard form $(a+b)^n$:

• The first term, $a = x^m$.
• The second term, $b = \frac{1}{x^2}$, which can be written as $x^{-2}$ for easier calculation with exponents.
• The power of the binomial, $n = 22$.

Explanation: Breaking down the given expression into its $a$, $b$, and $n$ components is the first crucial step. This allows us to correctly substitute these values into the general term formula. Rewriting $\frac{1}{x^2}$ as $x^{-2}$ simplifies the subsequent exponent arithmetic.

Step 2: Write the general term ($T_{r+1}$) for the expansion

Substitute $a$, $b$, and $n$ into the general term formula $T_{r+1} = {^nC_r} a^{n-r} b^r$: $T_{r+1} = {^{22}C_r} ({x^m})^{22-r} ({x^{-2}})^r$

Explanation: This step generates an algebraic expression for any term in the expansion. The variable $r$ (ranging from $0$ to $n$) determines which specific term we are considering.

Step 3: Simplify the powers of $x$ in the general term

Apply the exponent rules $(p^q)^s = p^{qs}$ and $p^q \cdot p^s = p^{q+s}$: $T_{r+1} = {^{22}C_r} x^{m(22-r)} x^{-2r}$ $T_{r+1} = {^{22}C_r} x^{22m - mr - 2r}$

Explanation: To find the coefficient of a specific power of $x$ (in this case, $x^1$), we must combine all the $x$ terms into a single base with a single exponent. This simplification isolates the power of $x$ that we will later set equal to $1$.

Step 4: Determine possible values of 'r' from the given coefficient

The problem states that the coefficient of $x$ is $1540$. From our simplified general term, the coefficient part is ${^{22}C_r}$. So, we need to find the value(s) of $r$ such that ${^{22}C_r} = 1540$. Let's compute binomial coefficients for small values of $r$:

• ${^{22}C_0} = 1$
• ${^{22}C_1} = 22$
• ${^{22}C_2} = \frac{22 \times 21}{2 \times 1} = 231$
• ${^{22}C_3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 22 \times 7 \times 10 = 1540$ Thus, one possible value for $r$ is $3$. Recall the property of binomial coefficients: ${^nC_r} = {^nC_{n-r}}$. Using this property, if ${^{22}C_3} = 1540$, then ${^{22}C_{22-3}} = {^{22}C_{19}}$ must also be $1540$. Therefore, the possible values for $r$ are $3$ and $19$.

Explanation: The numerical value of the coefficient is provided, which allows us to find the specific index $r$ that corresponds to this coefficient. By systematically checking values or recognizing common binomial coefficient results, we find $r=3$. The symmetry property ${^nC_r} = {^nC_{n-r}}$ is a valuable shortcut to find the second possible value of $r$ without further extensive calculation.

Step 5: Equate the power of $x$ to $1$ and solve for $m$ in terms of $r$

We are looking for the coefficient of $x$ (which means $x^1$). Therefore, the exponent of $x$ in our general term must be equal to $1$. Set the exponent of $x$ from Step 3 equal to $1$: $22m - mr - 2r = 1$ Now, we need to solve this equation for $m$: $m(22-r) = 1 + 2r$ $m = \frac{1 + 2r}{22-r}$

Explanation: The problem asks for the coefficient of $x$, implying $x^1$. By setting the derived exponent of $x$ equal to $1$, we establish a relationship between $m$ and $r$. Rearranging this equation to express $m$ in terms of $r$ prepares us to test the possible values of $r$ found in the previous step.

Step 6: Substitute the possible values of 'r' and check the condition for 'm'

The problem states that $m$ is a natural number (i.e., $m \in \{1, 2, 3, \dots\}$). We will test both values of $r$ we found:

Case 1: When $r=3$ Substitute $r=3$ into the equation for $m$: $m = \frac{1 + 2(3)}{22-3}$ $m = \frac{1 + 6}{19}$ $m = \frac{7}{19}$ Since $m = \frac{7}{19}$ is not a natural number, this value of $r$ is not valid.

Case 2: When $r=19$ Substitute $r=19$ into the equation for $m$: $m = \frac{1 + 2(19)}{22-19}$ $m = \frac{1 + 38}{3}$ $m = \frac{39}{3}$ $m = 13$ Since $m = 13$ is a natural number, this is the correct value for $m$.

Explanation: We evaluate $m$ for each valid $r$. It is crucial to remember and apply all constraints given in the problem statement, such as $m$ being a natural number. Any solution for $m$ that does not satisfy these constraints must be discarded.

Final Answer: The natural number $m$, for which the coefficient of $x$ in the binomial expansion of ${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$ is $1540$, is $\boxed{13}$.

Tips and Common Mistakes to Avoid:

• Careful with Exponent Simplification: A frequent error is mismanaging negative exponents (e.g., treating $(x^{-2})^r$ as $x^{2r}$ instead of $x^{-2r}$) or incorrectly combining powers (e.g., $x^p \cdot x^q \ne x^{p \cdot q}$). Always double-check your exponent rules.
• Understanding "Coefficient of $x$": The phrase "coefficient of $x$" specifically means the coefficient of $x^1$. If the question asked for the "constant term," the exponent would be $x^0$.
• Checking Constraints: Always verify that your final answer satisfies all conditions given in the problem, such as variables being natural numbers, integers, or falling within a specific range. Discard any solutions that do not meet these criteria.
• Symmetry of Binomial Coefficients: Remember that ${^nC_r} = {^nC_{n-r}}$. This property can significantly reduce computation time when finding values of $r$ from a given coefficient.

Summary/Key Takeaway: This problem is an excellent application of the Binomial Theorem, demonstrating how to find an unknown exponent ($m$) within the terms of a binomial expansion. The solution requires a methodical approach: setting up the general term, simplifying exponents, solving for the index $r$ using the given coefficient, and finally using the desired power of $x$ to solve for $m$. Always ensure the final answer adheres to any specified domain (e.g., natural numbers).