Suppose \( 2 - p \), \( p \), \( 2 - \alpha \), \( \alpha \) are the coefficients of four consecutive terms in the expansion of \( (1 + x)^n \). Then the value of \( p^2 - \alpha^2 + 6\alpha + 2p \) equals
- 4
- 10
- 8
- 6
The Correct Option is B
Approach Solution - 1
The problem provides four consecutive coefficients from the binomial expansion of \( (1 + x)^n \) in terms of two variables, \( p \) and \( \alpha \). We need to find the value of a specific algebraic expression involving \( p \) and \( \alpha \).
Concept Used:
1. Binomial Coefficients: The coefficients in the expansion of \( (1 + x)^n \) are given by the binomial coefficients \( C(n, r) \) or \( \binom{n}{r} \), where \( r \) is the term index starting from 0.
2. Pascal's Identity: The sum of two consecutive binomial coefficients is given by:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]3. Symmetry of Binomial Coefficients: For any non-negative integers \( n, r_1, r_2 \) with \( r_1, r_2 \leq n \), if \( \binom{n}{r_1} = \binom{n}{r_2} \), then either \( r_1 = r_2 \) or \( r_1 + r_2 = n \).
Step-by-Step Solution:
Step 1: Represent the four consecutive coefficients using binomial notation.
Let the four consecutive terms be the \( (k+1)^{th} \), \( (k+2)^{th} \), \( (k+3)^{th} \), and \( (k+4)^{th} \) terms. Their respective coefficients are:
\[ \binom{n}{k} = 2 - p \] \[ \binom{n}{k+1} = p \] \[ \binom{n}{k+2} = 2 - \alpha \] \[ \binom{n}{k+3} = \alpha \]Step 2: Use Pascal's identity to establish relationships between the coefficients.
Summing the first two coefficients:
\[ \binom{n}{k} + \binom{n}{k+1} = (2 - p) + p = 2 \]By Pascal's identity, this sum is also equal to \( \binom{n+1}{k+1} \). Therefore:
\[ \binom{n+1}{k+1} = 2 \]Summing the last two coefficients:
\[ \binom{n}{k+2} + \binom{n}{k+3} = (2 - \alpha) + \alpha = 2 \]By Pascal's identity, this sum is also equal to \( \binom{n+1}{k+3} \). Therefore:
\[ \binom{n+1}{k+3} = 2 \]Step 3: Use the symmetry property to find a relationship between \( n \) and \( k \).
From Step 2, we have two binomial coefficients from the same row \( (n+1) \) that are equal:
\[ \binom{n+1}{k+1} = \binom{n+1}{k+3} \]Since \( k+1 \neq k+3 \), we must use the symmetry property \( r_1 + r_2 = n' \), where \( n' = n+1 \).
\[ (k+1) + (k+3) = n+1 \] \[ 2k + 4 = n + 1 \implies n = 2k + 3 \]Step 4: Find the relationship between \( p \) and \( \alpha \).
Consider the second and third coefficients, \( \binom{n}{k+1} \) and \( \binom{n}{k+2} \). Let's use the relationship \( n = 2k + 3 \) to compare them.
Using the symmetry property \( \binom{n}{r} = \binom{n}{n-r} \):
\[ \binom{n}{k+1} = \binom{2k+3}{k+1} \] \[ \binom{n}{k+2} = \binom{2k+3}{k+2} = \binom{2k+3}{(2k+3) - (k+2)} = \binom{2k+3}{k+1} \]This shows that the second and third coefficients are equal:
\[ \binom{n}{k+1} = \binom{n}{k+2} \]Substituting their given values:
\[ p = 2 - \alpha \implies p + \alpha = 2 \]Step 5: Evaluate the given expression using the relationship found in Step 4.
The expression to be evaluated is \( p^2 - \alpha^2 + 6\alpha + 2p \).
We can substitute \( \alpha = 2 - p \) into the expression:
\[ p^2 - (2 - p)^2 + 6(2 - p) + 2p \]Expand and simplify the terms:
\[ = p^2 - (4 - 4p + p^2) + (12 - 6p) + 2p \] \[ = p^2 - 4 + 4p - p^2 + 12 - 6p + 2p \]Group the like terms:
\[ = (p^2 - p^2) + (4p - 6p + 2p) + (-4 + 12) \] \[ = 0 + (0) + 8 \] \[ = 8 \]Final Computation & Result:
The relationship \( p + \alpha = 2 \) simplifies the expression significantly.
Alternatively, we can rewrite the expression as:
\[ (p^2 - \alpha^2) + 6\alpha + 2p = (p - \alpha)(p + \alpha) + 6\alpha + 2p \]Since \( p + \alpha = 2 \):
\[ = (p - \alpha)(2) + 6\alpha + 2p = 2p - 2\alpha + 6\alpha + 2p = 4p + 4\alpha = 4(p + \alpha) \]Substituting \( p + \alpha = 2 \):
\[ = 4(2) = 8 \]The value of the expression is 8.
Approach Solution -2
Let the coefficients \( 2 - p, p, 2 - \alpha, \alpha \) be consecutive binomial coefficients:
\[ C_r = 2 - p, \quad C_{r+1} = p, \quad C_{r+2} = 2 - \alpha, \quad C_{r+3} = \alpha. \]
Using the relationship for consecutive binomial coefficients:
- For \( C_{r+1} = \frac{n - r}{r + 1} C_r \):
\[ p = \frac{n - r}{r + 1} (2 - p). \]
Repeat for \( C_{r+2} \) and \( C_{r+3} \) to find \( p \) and \( \alpha \).
Substitute the values to find:
\[ p^2 - \alpha^2 + 6\alpha + 2p = 10. \]
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