Find the term independent of \( x \) in the expansion of \( (1 + x)^{n} (1 + \frac{1}{x})^{n} \).
Show Hint
Whenever you see \((1+x)^n(1+1/x)^n\), it is mathematically identical to \(\frac{(1+x)^{2n}}{x^n}\). This trick quickly converts a product into a single binomial expansion problem.
Step 1: Understanding the Concept:
The term "independent of $x$" refers to the term in the binomial expansion where the power of $x$ is zero (i.e., the coefficient of $x^0$). We can simplify the given algebraic expression before looking for this specific term. Step 2: Key Formula or Approach:
Simplify the expression by combining the terms with the common exponent $n$:
\[ \text{Expression} = (1 + x)^n \left(1 + \frac{1}{x}\right)^n \]
Then use the general term of binomial expansion $T_{r+1} = ^NC_r a^{N-r} b^r$. Step 3: Detailed Explanation:
Let's find a common denominator for the second term:
\[ \left(1 + \frac{1}{x}\right) = \frac{x + 1}{x} \]
Substitute this back into the original expression:
\[ (1 + x)^n \left(\frac{x + 1}{x}\right)^n = (1 + x)^n \frac{(1 + x)^n}{x^n} \]
Combine the numerators using the rule $a^m \cdot a^n = a^{m+n}$:
\[ = \frac{(1 + x)^{2n}}{x^n} \]
We are looking for the term independent of $x$ in this entire fraction. Let the general term of $(1 + x)^{2n}$ be $T_{r+1}$.
\[ T_{r+1} \text{ in } \frac{(1 + x)^{2n}}{x^n} = \frac{^{2n}C_r x^r}{x^n} = ^{2n}C_r x^{r-n} \]
For the term to be independent of $x$, the exponent of $x$ must be $0$:
\[ r - n = 0 \implies r = n \]
Therefore, the coefficient of this term is obtained by substituting $r = n$:
\[ \text{Coefficient} = ^{2n}C_n = \binom{2n}{n} \] Step 4: Final Answer:
The term independent of $x$ is $\binom{2n}{n}$.