Key Concepts and Formulae

The binomial expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ The general term, often denoted as $T_{r+1}$, is $\binom{n}{r} a^{n-r} b^r$.

For the expansion of $(1+x)^n$, the general term is $T_{r+1} = \binom{n}{r} (1)^{n-r} x^r = \binom{n}{r} x^r$. The coefficient of $x^r$ in this expansion is $\binom{n}{r}$.

Finding the Middle Term(s):

• If $n$ is an even integer, there is only one middle term, which is the $(\frac{n}{2} + 1)^{th}$ term. Its coefficient is $\binom{n}{n/2}$.
• If $n$ is an odd integer, there are two middle terms: the $(\frac{n+1}{2})^{th}$ term and the $(\frac{n+3}{2})^{th}$ term. Their coefficients are $\binom{n}{(n-1)/2}$ and $\binom{n}{(n+1)/2}$ respectively.

Pascal's Identity: A crucial identity for binomial coefficients is Pascal's Identity: $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$ This identity simplifies the sum of two consecutive binomial coefficients.

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Step-by-Step Solution

1. Determine the coefficient of the middle term in the expansion of $(1+x)^{20}$.

• Explanation: Here, the power $n=20$, which is an even number. Therefore, there will be a single middle term.
• Calculation: The position of the middle term is $(\frac{n}{2} + 1)^{th} = (\frac{20}{2} + 1)^{th} = (10 + 1)^{th} = 11^{th}$ term.
• Result: The coefficient of the $11^{th}$ term ($T_{11}$) in $(1+x)^{20}$ is $\binom{20}{10}$. Let this be $C_1 = \binom{20}{10}$.

2. Determine the sum of the coefficients of the two middle terms in the expansion of $(1+x)^{19}$.

• Explanation: Here, the power $n=19$, which is an odd number. Therefore, there will be two middle terms.
• Calculation:
  • The position of the first middle term is $(\frac{n+1}{2})^{th} = (\frac{19+1}{2})^{th} = (\frac{20}{2})^{th} = 10^{th}$ term. Its coefficient is $\binom{19}{9}$ (since for $T_{r+1}$, $r=9$).
  • The position of the second middle term is $(\frac{n+3}{2})^{th} = (\frac{19+3}{2})^{th} = (\frac{22}{2})^{th} = 11^{th}$ term. Its coefficient is $\binom{19}{10}$ (since for $T_{r+1}$, $r=10$).
• Result: The sum of the coefficients of the two middle terms is $\binom{19}{9} + \binom{19}{10}$. Let this be $C_2 = \binom{19}{9} + \binom{19}{10}$.

3. Simplify the sum of coefficients using Pascal's Identity.

• Explanation: We can simplify $C_2$ using Pascal's Identity, $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.
• Application: In our case, $n=19$, $r=9$. So, $\binom{19}{9} + \binom{19}{10} = \binom{19+1}{9+1} = \binom{20}{10}$.
• Result: Therefore, $C_2 = \binom{20}{10}$.

4. Calculate the required ratio.

• Explanation: The question asks for the ratio of the coefficient of the middle term in $(1+x)^{20}$ to the sum of the coefficients of the two middle terms in $(1+x)^{19}$.
• Calculation: Required Ratio $= \frac{C_1}{C_2} = \frac{\binom{20}{10}}{\binom{20}{10}}$
• Result: Required Ratio $= 1$.

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Tips and Common Mistakes to Avoid

• Careful with 'n' for middle terms: Always remember to check if 'n' is even or odd to correctly identify the number and position of middle terms. A common mistake is to assume there's always one middle term or to incorrectly calculate its index.
• Understanding Binomial Coefficient Notation: $\binom{n}{r}$ (or $^{n}C_r$) represents the coefficient of the $(r+1)^{th}$ term when expanding $(a+b)^n$. If you are looking for the coefficient of the $k^{th}$ term, it will be $\binom{n}{k-1}$.
• Pascal's Identity is your friend: Recognize opportunities to use identities like Pascal's Identity ($\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$) to simplify expressions, especially in objective-type questions where speed is important.

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Summary

The problem efficiently tests the understanding of finding middle terms in binomial expansions and the application of Pascal's Identity. By correctly identifying the coefficient of the single middle term for an even power and the sum of the coefficients of the two middle terms for an odd power, and then applying Pascal's Identity, the problem simplifies to a straightforward division of identical terms, resulting in a ratio of 1.