If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$, then the sum of the product of the coefficients taken two at a time can be represented by $\displaystyle\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j = 2^a - \dfrac{b!}{c(d!)^2}$. Then which of the following are correct?
Step-by-Step Solution
Key Concept: The key idea is to relate the sum of products of coefficients taken two at a time, $\sum_{i<j} C_i C_j$, to the square of the sum of all coefficients, $(\sum C_i)^2$, using the identity $(\sum C_i)^2 = \sum C_i^2 + 2 \sum_{i<j} C_i C_j$, and then substitute the standard results for $\sum C_i$ and $\sum C_i^2$.
Step 1: Use the identity $(C_0 + C_1 + \cdots + C_n)^2 = C_0^2 + C_1^2 + \cdots + C_n^2 + 2\sum_{i<j} C_i C_j$. We know $\sum C_i = 2^n$, so $(2^n)^2 = 2^{2n}$.
Step 2: Also, $\sum_{i=0}^n C_i^2 = \binom{2n}{n} = \dfrac{(2n)!}{(n!)^2}$ (by Vandermonde's identity).
Step 3: Therefore:
$$2\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j = 2^{2n} - \frac{(2n)!}{(n!)^2}$$
$$\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j = \frac{2^{2n} - \frac{(2n)!}{(n!)^2}}{2} = 2^{2n-1} - \frac{(2n)!}{2(n!)^2}$$
Step 4: Comparing with $2^a - \dfrac{b!}{c(d!)^2}$:
$$a = 2n-1,\quad b = 2n,\quad c = 2,\quad d = n$$
Step 5: All four options (a), (b), (c), (d) are correct.
Correct Answer: 1, 2, 3, 4