Key Concept: Properties of Binomial Coefficients

This problem utilizes the fundamental properties of binomial coefficients, specifically the ratio between consecutive coefficients in the expansion of $(1+x)^n$. The coefficient of the $(k+1)^{th}$ term in the expansion of $(1+x)^n$ is given by ${}^n C_k$. A crucial identity for solving such problems is the ratio of consecutive binomial coefficients:

$\frac{{}^n C_k}{{}^n C_{k-1}} = \frac{n-k+1}{k}$

This formula allows us to establish relationships between $n$ and $r$ (the term index) when ratios of consecutive coefficients are given.

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Step-by-Step Derivations

1. Identifying the Coefficients and Setting up Ratios: Let the three consecutive terms in the expansion of $(1+x)^n$ have coefficients ${}^n C_{r-1}$, ${}^n C_r$, and ${}^n C_{r+1}$. These represent the coefficients of the $r^{th}$, $(r+1)^{th}$, and $(r+2)^{th}$ terms, respectively.

We are given that their ratio is $1:5:20$. This means: $\frac{{}^n C_{r-1}}{{}^n C_r} = \frac{1}{5} \quad \text{and} \quad \frac{{}^n C_r}{{}^n C_{r+1}} = \frac{5}{20} = \frac{1}{4}$

2. Solving the First Ratio: Consider the first ratio: $\frac{{}^n C_{r-1}}{{}^n C_r} = \frac{1}{5}$ To use our standard ratio formula $\frac{{}^n C_k}{{}^n C_{k-1}} = \frac{n-k+1}{k}$, we can invert this ratio. Here, our $k$ corresponds to $r$. $\frac{{}^n C_r}{{}^n C_{r-1}} = 5$ Applying the formula with $k=r$: $\frac{n-r+1}{r} = 5$ To eliminate the denominator, we multiply both sides by $r$: $n-r+1 = 5r$ Rearranging the terms to express $n$ in terms of $r$: $n = 5r + r - 1$ $n = 6r - 1 \quad \text{(Equation 1)}$ Explanation: This step establishes the first relationship between the unknown variables $n$ and $r$ using the given ratio of the first two consecutive coefficients.

3. Solving the Second Ratio: Now, consider the second ratio: $\frac{{}^n C_r}{{}^n C_{r+1}} = \frac{1}{4}$ Again, to align with our ratio formula $\frac{{}^n C_k}{{}^n C_{k-1}} = \frac{n-k+1}{k}$, we can consider the reciprocal. Here, if we take $k$ as $r+1$, then $k-1$ is $r$. $\frac{{}^n C_{r+1}}{{}^n C_r} = 4$ Applying the formula with $k=(r+1)$: $\frac{n-(r+1)+1}{r+1} = 4$ Simplify the numerator: $\frac{n-r-1+1}{r+1} = 4$ $\frac{n-r}{r+1} = 4$ Multiply both sides by $(r+1)$: $n-r = 4(r+1)$ $n-r = 4r+4$ Rearranging the terms to express $n$ in terms of $r$: $n = 4r + r + 4$ $n = 5r + 4 \quad \text{(Equation 2)}$ Explanation: This step provides the second relationship between $n$ and $r$, derived from the ratio of the next pair of consecutive coefficients.

4. Determining the Values of $n$ and $r$: We now have a system of two linear equations with two variables:

1. $n = 6r - 1$
2. $n = 5r + 4$

Since both equations are equal to $n$, we can set them equal to each other: $6r - 1 = 5r + 4$ To solve for $r$, gather the $r$ terms on one side and constants on the other: $6r - 5r = 4 + 1$ $r = 5$ Now substitute the value of $r=5$ into either Equation 1 or Equation 2 to find $n$. Using Equation 2: $n = 5(5) + 4$ $n = 25 + 4$ $n = 29$ Explanation: By solving the system of equations, we precisely determine the values of $n$ (the power of the binomial expansion) and $r$ (the starting index of the consecutive terms).

5. Calculating the Coefficient of the Fourth Term: The problem asks for the coefficient of the fourth term. In the binomial expansion $(1+x)^n$, the $(k+1)^{th}$ term is $T_{k+1}$, and its coefficient is ${}^n C_k$. For the fourth term, $T_4$, we set $k+1 = 4$, which means $k=3$. So, the coefficient of the fourth term is ${}^n C_3$.

Substitute the values of $n=29$ and $k=3$: Coefficient of the fourth term = ${}^{29} C_3$ Recall the formula for combinations: ${}^n C_k = \frac{n!}{k!(n-k)!}$. ${}^{29} C_3 = \frac{29!}{3!(29-3)!} = \frac{29!}{3!26!}$ Expand the factorials: ${}^{29} C_3 = \frac{29 \times 28 \times 27 \times 26!}{ (3 \times 2 \times 1) \times 26!}$ Cancel out $26!$ from the numerator and denominator: ${}^{29} C_3 = \frac{29 \times 28 \times 27}{3 \times 2 \times 1}$ Perform the division to simplify calculations: ${}^{29} C_3 = 29 \times \left(\frac{28}{2}\right) \times \left(\frac{27}{3}\right)$ ${}^{29} C_3 = 29 \times 14 \times 9$ Now, multiply the numbers: ${}^{29} C_3 = 29 \times 126$ ${}^{29} C_3 = 3654$ Explanation: This final step applies the definition of binomial coefficients using the determined values of $n$ to calculate the specific coefficient requested by the problem.

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Tips and Common Mistakes:

• Indexing: Be very careful with the indexing of binomial coefficients. If the problem refers to the $r^{th}$ term, its coefficient is ${}^n C_{r-1}$. If it refers to $C_r$ as the $r^{th}$ coefficient, it means the coefficient of $x^r$, which is ${}^n C_r$ (i.e., the $(r+1)^{th}$ term). The solution assumes the coefficients are ${}^n C_{r-1}$, ${}^n C_r$, and ${}^n C_{r+1}$ for three consecutive terms.
• Ratio Formula Application: Double-check whether you are using $\frac{{}^n C_k}{{}^n C_{k-1}}$ or $\frac{{}^n C_{k-1}}{{}^n C_k}$. Inverting the ratio or swapping $k$ and $k-1$ in the formula can lead to errors.
• Algebraic Precision: Ensure careful algebraic manipulation when solving the simultaneous equations for $n$ and $r$. A small error can propagate through the entire solution.
• Calculation of Combinations: When calculating ${}^n C_k$, simplify the expression by canceling terms before multiplying large numbers to avoid errors.

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Summary:

We successfully determined the values of $n$ and $r$ by setting up and solving a system of equations derived from the given ratios of consecutive binomial coefficients using the identity $\frac{{}^n C_k}{{}^n C_{k-1}} = \frac{n-k+1}{k}$. Once $n=29$ was found, the coefficient of the fourth term, ${}^{29} C_3$, was calculated to be $3654$. This problem highlights the importance of understanding and correctly applying the properties of binomial coefficients.

The final answer is $\boxed{\text{3654}}$.