Question

If the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1:5:20, then the coefficient of the fourth term of the expansion is?

• A. 3654
• B. 3658
• C. 3600
• D. 1000

Hint:

Use the properties of binomial coefficients and ratios to determine terms in expansions efficiently.

Answer 1

The Correct Option is A

Step 1: Define the consecutive terms.
- Let the three consecutive terms be \(\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}\).
- The given ratio is:
\[ \binom{n}{r-1} : \binom{n}{r} : \binom{n}{r+1} = 1 : 5 : 20. \]
Step 2: Express the ratios.
- From the ratio \(\frac{\binom{n}{r}}{\binom{n}{r-1}} = 5\):
\[ \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}} = 5 \implies \frac{(n-r+1)}{r} = 5. \]
\[ n - r + 1 = 5r \implies n = 6r - 1. \]
- From the ratio \(\frac{\binom{n}{r+1}}{\binom{n}{r}} = 4\):
\[ \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{r!(n-r)!}} = 4 \implies \frac{(n-r)}{r+1} = 4. \]
\[ n - r = 4r + 4 \implies n = 5r + 4. \]
Step 3: Solve for \(r\) and \(n\).
- Equating the two expressions for \(n\):
\[ 6r - 1 = 5r + 4 \implies r = 5, \quad n = 6(5) - 1 = 29. \]
Step 4: Find the coefficient of the fourth term.
- The fourth term corresponds to \(r = 3\):
\[ \binom{29}{3} = \frac{29 \times 28 \times 27}{3 \times 2 \times 1} = 3654. \]
Final Answer: The coefficient of the fourth term is \(3654\).