Key Concepts: The Binomial Theorem

The core concept for solving this problem is the Binomial Theorem, which provides a formula for expanding expressions of the form $(a+b)^n$. The general $(r+1)^{th}$ term (denoted as $T_{r+1}$) in the binomial expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^n C_r a^{n-r} b^r$ For the specific binomial factor $(1-x)^n$ in our problem, we set $a=1$ and $b=-x$. Substituting these into the general term formula yields: $T_{r+1} = {}^n C_r (1)^{n-r} (-x)^r = {}^n C_r (-1)^r x^r$ This general term allows us to find the coefficient of any specific power of $x$ within the expansion of $(1-x)^n$.

Step-by-Step Derivation

Our objective is to determine the coefficient of ${x^n}$ in the expansion of ${(1+x)(1-x)^n}$.

Step 1: Decomposing the Expression To simplify the process, we first expand the given expression by distributing $(1+x)$. This is a crucial first step as it breaks down a product into a sum of terms, which are easier to handle individually. $(1+x)(1-x)^n = (1) \cdot (1-x)^n + (x) \cdot (1-x)^n$ $= (1-x)^n + x(1-x)^n$ Now, we need to find the coefficient of ${x^n}$ in each of these two parts and then sum them up.

Step 2: Finding the Coefficient of $x^n$ in the First Part, $(1-x)^n$ For the expression $(1-x)^n$, we use its general term ${T_{r+1} = {}^n C_r (-1)^r x^r}$. To find the coefficient of ${x^n}$, we need the power of $x$ to be $n$. Therefore, we set $r=n$. Substituting $r=n$ into the general term, we get: $T_{n+1} = {}^n C_n (-1)^n x^n$ The coefficient of ${x^n}$ in $(1-x)^n$ is ${{}^n C_n (-1)^n}$. Since ${{}^n C_n = 1}$ (there is only one way to choose all $n$ items from $n$), this coefficient simplifies to ${(-1)^n}$.

Step 3: Finding the Coefficient of $x^n$ in the Second Part, $x(1-x)^n$ For the expression $x(1-x)^n$, to obtain a term containing ${x^n}$, we must multiply the external $x$ by a term containing ${x^{n-1}}$ from the expansion of $(1-x)^n$. Thus, we need to find the coefficient of ${x^{n-1}}$ in $(1-x)^n$. Using the general term formula ${T_{r+1} = {}^n C_r (-1)^r x^r}$ for $(1-x)^n$, we set the power of $x$ to $n-1$, which means $r=n-1$. The term containing ${x^{n-1}}$ is: $T_{(n-1)+1} = T_n = {}^n C_{n-1} (-1)^{n-1} x^{n-1}$ The coefficient of ${x^{n-1}}$ in $(1-x)^n$ is ${{}^n C_{n-1} (-1)^{n-1}}$. We use the property of binomial coefficients that ${{}^n C_r = {}^n C_{n-r}}$. Therefore, ${{}^n C_{n-1} = {}^n C_1 = n}$. So, the coefficient of ${x^{n-1}}$ in $(1-x)^n$ is ${n (-1)^{n-1}}$. When this coefficient is multiplied by the external $x$, it gives the coefficient of ${x^n}$ in the expression $x(1-x)^n$, which is also ${n (-1)^{n-1}}$.

Step 4: Combining the Coefficients The total coefficient of ${x^n}$ in the original expansion is the sum of the coefficients found in Step 2 and Step 3: Total Coefficient ($C$) = (Coefficient from $(1-x)^n$) + (Coefficient from $x(1-x)^n$) $C = (-1)^n + n(-1)^{n-1}$ Now, we simplify this expression. We observe that ${(-1)^n = (-1)^1 \cdot (-1)^{n-1} = -1 \cdot (-1)^{n-1}}$. We can factor out the common term ${(-1)^{n-1}}$: $C = (-1) \cdot (-1)^{n-1} + n \cdot (-1)^{n-1}$ $C = (-1)^{n-1} (-1 + n)$ $C = (-1)^{n-1} (n-1)$ This expression can also be written in an alternative form by manipulating the power of -1: $C = (-1)^{n-1} (n-1) = (-1)^n \cdot (-1)^{-1} \cdot (n-1) = (-1)^n \cdot (-1) \cdot (n-1)$ $C = (-1)^n (-(n-1)) = (-1)^n (1-n)$

Thus, the coefficient of ${x^n}$ in the expansion of ${(1+x)(1-x)^n}$ is ${(-1)^{n-1}(n-1)}$ or equivalently ${(-1)^n(1-n)}$.

Tips and Common Mistakes

• Sign Convention: Always be careful with the alternating signs arising from the ${(-1)^r}$ term, especially when dealing with $(1-x)^n$. A common error is to miss or incorrectly apply this sign.
• Index Management: When you have a factor like $x$ multiplying a binomial expansion, remember that you are looking for a term with a lower power of $x$ (e.g., $x^{n-1}$ for $x(1-x)^n$) within the expansion itself.
• Binomial Coefficient Properties: Utilize properties like ${{}^n C_r = {}^n C_{n-r}}$ to simplify binomial coefficients. For example, ${{}^n C_{n-1}}$ is more easily recognized as $n$ by converting it to ${{}^n C_1}$.
• Algebraic Simplification: Practice factoring out common terms, especially powers of ${(-1)}$, to arrive at the simplest and most elegant form of the answer.

Summary and Key Takeaway

This problem is a classic application of the Binomial Theorem for finding coefficients in composite expressions. The key strategy involves:

1. Decomposing the complex expression into a sum of simpler terms.
2. Independently finding the coefficient for the desired power of $x$ in each simpler term.
3. Summing these individual coefficients.
4. Carefully simplifying the final algebraic expression. Following these steps, we precisely determined that the coefficient of ${x^n}$ in the expansion of ${(1+x)(1-x)^n}$ is ${(-1)^{n-1}(n-1)}$.