If for `n in I , n > 10 ;1+(1+x)+(1+x)^2++(1+x)^n=sum_(k=0)^n a_k*x^k , x!=0` then

Options

A. `a_(n-2)=(n(n+1))/(2)`
B. `a_(9)^(2)-a_(8)^(2)=^(n+2)C_(10)('^(n+1)C_(10)-"^(n+1)C_(9))`
C. `a_(p) gt a_(p-1)` for `p lt (n)/(2)`
D. `sum_(k=0)^(n)a_(k)=2^(n+1)`

Explanation

To solve the given problem, we need to analyze the expression and find the coefficients of the polynomial expansion. Let's break it down step by step.

### Step 1: Understanding the Expression
We are given the expression:
\[ S = 1 + (1+x) + (1+x)^2 + \ldots + (1+x)^n \]
This is a sum of terms from \( (1+x)^0 \) to \( (1+x)^n \).

### Step 2: Recognizing the Series
This series can be recognized as a geometric series. The first term \( a = 1 \) and the common ratio \( r = (1+x) \). The number of terms is \( n + 1 \).

### Step 3: Using the Geometric Series Formula
The sum of a geometric series can be calculated using the formula:
\[ S = a \frac{r^{n+1} - 1}{r - 1} \]
Substituting the values we have:
\[ S = 1 \cdot \frac{(1+x)^{n+1} - 1}{(1+x) - 1} = \frac{(1+x)^{n+1} - 1}{x} \]

### Step 4: Expressing the Sum
Thus, we can express the sum as:
\[ S = \frac{(1+x)^{n+1} - 1}{x} \]

### Step 5: Finding Coefficients
We can now express \( S \) as:
\[ S = \sum_{k=0}^{n} a_k x^k \]
To find \( a_k \), we need to analyze the expression \( (1+x)^{n+1} \).

### Step 6: Coefficient Extraction
The coefficients \( a_k \) can be found using the binomial expansion:
\[ (1+x)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} x^k \]
Thus, the coefficients \( a_k \) can be expressed as:
\[ a_k = \binom{n+1}{k} \]

### Step 7: Evaluating Specific Coefficients
To evaluate specific coefficients, we can use:
- \( a_0 = \binom{n+1}{0} = 1 \)
- \( a_1 = \binom{n+1}{1} = n+1 \)
- \( a_2 = \binom{n+1}{2} = \frac{(n+1)n}{2} \)

### Step 8: Conclusion
From the analysis, we can conclude that:
- The coefficients \( a_k \) are given by \( a_k = \binom{n+1}{k} \).
- The conditions provided in the problem can be verified based on these coefficients.