Question

The sum of the coefficients of three consecutive terms in the binomial expansion of (1 + x)ⁿ⁺², which are in the ratio 1 : 3 : 5, is equal to

• A. 25
• B. 63
• C. 92
• D. 41

Answer 1

The Correct Option is B

We are tasked with solving the given ratios involving binomial coefficients and calculating the sum of specific combinations.

Step 1: Express the First Ratio

The first ratio is given as: \[ \frac{\binom{n+2}{r-1}}{\binom{n+2}{r}} = \frac{1}{3}. \]

Using the formula for binomial coefficients: \[ \binom{n+2}{r-1} = \frac{(n+2)!}{(r-1)!(n+3-r)!}, \quad \binom{n+2}{r} = \frac{(n+2)!}{r!(n+2-r)!}. \]

Simplify the ratio: \[ \frac{\binom{n+2}{r-1}}{\binom{n+2}{r}} = \frac{r}{n-r+3} = \frac{1}{3}. \]

Cross-multiply: \[ n - r + 3 = 3r \implies n = 4r - 3 \quad \text{...(1)}. \]

Step 2: Express the Second Ratio

The second ratio is given as: \[ \frac{\binom{n+2}{r}}{\binom{n+2}{r+1}} = \frac{3}{5}. \]

Using the binomial coefficient formula: \[ \binom{n+2}{r+1} = \frac{(n+2)!}{(r+1)!(n+1-r)!}. \]

Simplify the ratio: \[ \frac{\binom{n+2}{r}}{\binom{n+2}{r+1}} = \frac{r+1}{n+2-r} = \frac{3}{5}. \]

Cross-multiply: \[ 5(r+1) = 3(n+2-r) \implies 5r + 5 = 3n + 6 - 3r \implies 8r - 1 = 3n \quad \text{...(2)}. \]

Step 3: Solve Equations (1) and (2)

Substitute \( n = 4r - 3 \) from equation (1) into equation (2): \[ 8r - 1 = 3(4r - 3) \implies 8r - 1 = 12r - 9 \implies 4r = 8 \implies r = 2. \]

Step 4: Solve for \( n \)

Substitute \( r = 2 \) into equation (1): \[ n = 4(2) - 3 = 5. \]

Step 5: Calculate the Sum of Combinations

The sum is given by: \[ \binom{7}{1} + \binom{7}{2} + \binom{7}{3}. \]

Calculate each term: \[ \binom{7}{1} = 7, \quad \binom{7}{2} = 21, \quad \binom{7}{3} = 35. \]

Total sum: \[ 7 + 21 + 35 = 63. \]

Final Answer:

The sum of the combinations is 63.