Binomial Theorem Question - None

Question

Note that $C_k$ is binomial coefficient $^nC_k$. Consider $\displaystyle\left(\sum_{i=0}^{n} C_i\right)^2 = \sum_{i=0}^{n} C_i^2 + 2\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j$. Which of the following are correct?

(a) $\displaystyle\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j = 2^{2n-1} - \frac{(2n)!}{2(n!)^2}$

(b) The value of $\displaystyle\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j$ equals $2^{2n-1} - \frac{1}{2}\cdot {}^{2n}C_n$

(c) $a = 2n-1,\ b = 2n,\ c = 2,\ d = n$

(d) All of the above

Accepted Answer

Step 1: Start with $(\sum_{i=0}^{n} C_i)^2 = 4^n = (2^n)^2$

Step 2: Use Vandermonde's identity: $\sum_{i=0}^{n} C_i^2 = \sum_{i=0}^{n} \binom{n}{i}\binom{n}{n-i} = \binom{2n}{n}$

Step 3: From the given equation: $4^n = \binom{2n}{n} + 2\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j$

Step 4: Solve for the double sum: $\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j = \frac{4^n - \binom{2n}{n}}{2} = \frac{2^{2n} - \binom{2n}{n}}{2} = 2^{2n-1} - \frac{1}{2}\binom{2n}{n}$

Step 5: Note that $\binom{2n}{n} = \frac{(2n)!}{n! \cdot n!} = \frac{(2n)!}{(n!)^2}$, so $\frac{1}{2}\binom{2n}{n} = \frac{(2n)!}{2(n!)^2}$

Step 6: Therefore both (a) and (b) are correct (they are equivalent expressions)

Step 7: Statement (c) refers to parameters in a general formula, which is satisfied by the derivation above

∴ Answer: a,b,c,d

Question:

Step-by-Step Solution

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