Key Concept: Properties of Binomial Coefficients and Ratios of Consecutive Terms

In the binomial expansion of $(1+x)^n$, the coefficient of the $(k+1)^{th}$ term is given by $\binom{n}{k}$. For four consecutive terms, let their general indices be $r, r+1, r+2, r+3$. The coefficients are $\binom{n}{r}$, $\binom{n}{r+1}$, $\binom{n}{r+2}$, and $\binom{n}{r+3}$ respectively.

A crucial property of binomial coefficients is the ratio of consecutive terms: $\frac{\text{Coefficient of } T_{k+1}}{\text{Coefficient of } T_k} = \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$ Applying this to our sequence of terms:

• The ratio of $\binom{n}{r+1}$ to $\binom{n}{r}$ is $\frac{n-r}{r+1}$.
• The ratio of $\binom{n}{r+2}$ to $\binom{n}{r+1}$ is $\frac{n-(r+1)}{r+2} = \frac{n-r-1}{r+2}$.
• The ratio of $\binom{n}{r+3}$ to $\binom{n}{r+2}$ is $\frac{n-(r+2)}{r+3} = \frac{n-r-2}{r+3}$.

Step-by-step Derivation

1. Set Up Equations Based on Given Coefficients: We are given that the coefficients of four consecutive terms are $2-p$, $p$, $2-\alpha$, and $\alpha$. Let these correspond to $\binom{n}{r}$, $\binom{n}{r+1}$, $\binom{n}{r+2}$, and $\binom{n}{r+3}$ respectively. $\binom{n}{r} = 2-p \quad (i)$ $\binom{n}{r+1} = p \quad (ii)$ $\binom{n}{r+2} = 2-\alpha \quad (iii)$ $\binom{n}{r+3} = \alpha \quad (iv)$ Explanation: We equate the given expressions to the standard binomial coefficients for consecutive terms. This is the foundation for establishing relationships between $p, \alpha, n,$ and $r$.

2. Derive Initial Relationships for $p$ and $\alpha$: Using the ratio property for equations $(i)$ and $(ii)$: $\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{p}{2-p} = \frac{n-r}{r+1}$ Cross-multiplying and rearranging terms: $p(r+1) = (2-p)(n-r)$ $pr + p = 2n - 2r - pn + pr$ $p(n+1) = 2(n-r) \quad \text{(Equation A)}$ Similarly, using the ratio property for equations $(iii)$ and $(iv)$: $\frac{\binom{n}{r+3}}{\binom{n}{r+2}} = \frac{\alpha}{2-\alpha} = \frac{n-r-2}{r+3}$ Cross-multiplying and rearranging terms: $\alpha(r+3) = (2-\alpha)(n-r-2)$ $\alpha r + 3\alpha = 2n - 2r - 4 - \alpha n + \alpha r + 2\alpha$ $\alpha(n+1) = 2n - 2r - 4 \quad \text{(Equation B)}$ Explanation: By applying the ratio formula and simplifying, we obtain two key equations relating $p$, $\alpha$, $n$, and $r$. These equations form the basis for further deductions.

3. Find a Relationship Between $p$ and $\alpha$: Subtract Equation B from Equation A: $p(n+1) - \alpha(n+1) = (2n-2r) - (2n-2r-4)$ $(p-\alpha)(n+1) = 4 \quad \text{(Equation C)}$ Explanation: This step elegantly provides a direct relationship for $p-\alpha$ in terms of $n$, highlighting how these variables are linked. This will be useful when evaluating the final expression.

4. Determine the Relationship Between $n$ and $r$: Now, let's use the ratio property for equations $(ii)$ and $(iii)$: $\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{2-\alpha}{p} = \frac{n-r-1}{r+2} \quad \text{(Equation D)}$ From Equation A, we have $p = \frac{2(n-r)}{n+1}$. From Equation B, we can express $2-\alpha$: $2-\alpha = 2 - \frac{2(n-r-2)}{n+1} = \frac{2(n+1) - 2(n-r-2)}{n+1} = \frac{2n+2-2n+2r+4}{n+1} = \frac{2r+6}{n+1}$ Substitute these expressions for $p$ and $2-\alpha$ into Equation D: $\frac{\frac{2r+6}{n+1}}{\frac{2(n-r)}{n+1}} = \frac{n-r-1}{r+2}$ $\frac{2(r+3)}{2(n-r)} = \frac{n-r-1}{r+2}$ $\frac{r+3}{n-r} = \frac{n-r-1}{r+2}$ Cross-multiply: $(r+3)(r+2) = (n-r)(n-r-1)$ Since $(r+3)$ and $(r+2)$ are consecutive integers, and $(n-r)$ and $(n-r-1)$ are also consecutive integers, for this equality to hold, we must have: $n-r = r+3 \quad \text{and} \quad n-r-1 = r+2$ Both conditions consistently lead to the same relationship: $n = 2r+3 \quad \text{(Equation E)}$ Explanation: This is a crucial step. By combining the second ratio with the expressions for $p$ and $2-\alpha$ (derived from Equations A and B), we derive a polynomial identity. Comparing the consecutive terms in the product reveals a direct algebraic relationship between $n$ and $r$. This relationship is vital for simplifying later expressions.

5. Simplify $p+\alpha$: From Equation A: $p(n+1) = 2(n-r)$ From Equation B: $\alpha(n+1) = 2(n-r-2)$ Add these two equations together: $p(n+1) + \alpha(n+1) = 2(n-r) + 2(n-r-2)$ $(p+\alpha)(n+1) = 2n-2r+2n-2r-4$ $(p+\alpha)(n+1) = 4n-4r-4$ $(p+\alpha)(n+1) = 4(n-r-1) \quad \text{(Equation F)}$ Now, substitute the relationship $n=2r+3$ (from Equation E) into Equation F: $(p+\alpha)((2r+3)+1) = 4((2r+3)-r-1)$ $(p+\alpha)(2r+4) = 4(r+2)$ $(p+\alpha)2(r+2) = 4(r+2)$ Since $r$ is a non-negative integer (as it's an index for binomial coefficients), $r+2 \neq 0$. We can divide both sides by $2(r+2)$: $p+\alpha = 2 \quad \text{(Equation G)}$ Explanation: We sum the equations for $p(n+1)$ and $\alpha(n+1)$ to find an expression for $p+\alpha$. Substituting the relationship $n=2r+3$ derived earlier leads to a remarkable simplification, revealing that $p+\alpha$ has a constant value. This constant value is key to solving the problem.

6. Evaluate the Given Expression: We need to find the value of $p^2 - \alpha^2 + 6\alpha + 2p$. First, we use the difference of squares identity: $p^2 - \alpha^2 = (p-\alpha)(p+\alpha)$. So the expression becomes: $(p-\alpha)(p+\alpha) + 6\alpha + 2p$ From Equation G, we know that $p+\alpha=2$. Substitute this into the expression: $(p-\alpha)(2) + 6\alpha + 2p$ $= 2p - 2\alpha + 6\alpha + 2p$ Combine like terms: $= (2p+2p) + (-2\alpha+6\alpha)$ $= 4p + 4\alpha$ Factor out 4: $= 4(p+\alpha)$ Finally, substitute $p+\alpha=2$ (from Equation G) into this simplified expression: $= 4(2)$ $= 8$

Explanation: We leverage the difference of squares identity to factor $p^2 - \alpha^2$. By substituting the derived value of $p+\alpha=2$, the entire expression simplifies significantly, revealing its final constant value.

Final Answer The value of the expression $p^2 - \alpha^2 + 6\alpha + 2p$ is $\boxed{8}$.

Tips and Common Mistakes:

• Careful with Indices: Ensure you correctly map the terms $T_{r+1}, T_{r+2}, \dots$ to their corresponding binomial coefficients $\binom{n}{r}, \binom{n}{r+1}, \dots$. A common mistake is off-by-one errors in indices.
• Algebraic Precision: Binomial coefficient problems often involve extensive algebraic manipulation. Double-check each step, especially when cross-multiplying or combining terms.
• Recognize Key Identities: Pascal's identity and the ratio property of binomial coefficients are fundamental. Knowing when and how to apply them is crucial.
• Context of Variables: Remember that $n$ and $r$ (or $k$) must be non-negative integers for binomial coefficients $\binom{n}{r}$ to be defined in this context. While $p$ and $\alpha$ are variables describing the coefficients, the coefficients themselves must be integers.

Key Takeaway This problem demonstrates how to use the properties of consecutive binomial coefficients to establish a system of equations involving $n, r, p,$ and $\alpha$. Through careful algebraic manipulation and substitution, complex expressions can often be simplified to a constant value, even when the individual variables $n$ and $r$ are not explicitly determined. The consistency of mathematical derivation is paramount.