Question

The ratio of the coefficient of the middle term in the expansion of \((1+x)^{20}\) and the sum of the coefficients of two middle terms in expansion of \((1+x)^{19}\) is __________.

Hint:

The sum of the coefficients of the two middle terms of $(1+x)^{2n-1}$ is always equal to the coefficient of the single middle term of $(1+x)^{2n}$. This is a direct consequence of the recurrence relation of binomial coefficients.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
For \((1+x)^n\), if \(n\) is even, there is one middle term at position \(\frac{n}{2}+1\).
If \(n\) is odd, there are two middle terms at positions \(\frac{n+1}{2}\) and \(\frac{n+1}{2}+1\).
We use the property of binomial coefficients \(\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}\).
Step 2: Key Formula or Approach:
1. Coefficient of middle term of \((1+x)^{2m}\) is \(\binom{2m}{m}\).
2. Coefficients of two middle terms of \((1+x)^{2m-1}\) are \(\binom{2m-1}{m-1}\) and \(\binom{2m-1}{m}\).
Step 3: Detailed Explanation:
1. Expansion of \((1+x)^{20\):}
Here \(n=20\) (even). The middle term is the 11th term (\(r=10\)).
Coefficient \(C_1 = \binom{20}{10}\).

2. Expansion of \((1+x)^{19\):}
Here \(n=19\) (odd). The two middle terms are the 10th and 11th terms (\(r=9\) and \(r=10\)).
Sum of coefficients \(C_2 = \binom{19}{9} + \binom{19}{10}\).

Using Pascal's Identity \(\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}\):
\(C_2 = \binom{19}{10} + \binom{19}{9} = \binom{19+1}{10} = \binom{20}{10}\).

3. Calculate the ratio:
Ratio \(= \frac{C_1}{C_2} = \frac{\binom{20}{10}}{\binom{20}{10}} = 1\).
Step 4: Final Answer:
The ratio is 1.