Question

The value of $-{ }^{15} C _{1}+2 .{ }^{15} C _{2}-3 .{ }^{15} C _{3}+\ldots $ $-15 .{ }^{15} C _{15}+{ }^{14} C _{1}+{ }^{14} C _{3}+{ }^{14} C _{5}+\ldots .+{ }^{14} C _{11}$ is

• A. $2^{16}-1$
• B. $2^{13}-14$
• C. $2^{14}$
• D. $2^{13}-13$

Answer 1

The Correct Option is B

To solve the given problem, we need to evaluate the expression:

\[- { }^{15} C _{1}+2 .{ }^{15} C _{2}-3 .{ }^{15} C _{3}+\ldots -15 .{ }^{15} C _{15}+{ }^{14} C _{1}+{ }^{14} C _{3}+{ }^{14} C _{5}+\ldots +{ }^{14} C _{11}.\]

This expression is composed of two parts:

1. \(-{ }^{15} C_{1} + 2 \cdot { }^{15} C_{2} - 3 \cdot { }^{15} C_{3} + \ldots - 15 \cdot { }^{15} C_{15}\)
2. \({ }^{14} C_{1} + { }^{14} C_{3} + { }^{14} C_{5} + \ldots + { }^{14} C_{11}\)

1. Evaluate the First Part:

The first part can be represented as:

\(\sum_{k=1}^{15} (-1)^{k} \cdot k \cdot { }^{15} C_{k}\)

Now, consider the binomial expansion:

\((1-1)^{15} = \sum_{k=0}^{15} { }^{15} C_{k} \cdot (-1)^k = 0\)

Taking the derivative with respect to \(x\) in the expansion \((1 + x)^{15}\) and setting \(x = -1\) gives:

\((1 + x)^{15} = \sum_{k=0}^{15} { }^{15} C_{k} \cdot x^{k}\)

\(\Rightarrow \frac{d}{dx}\left((1 + x)^{15}\right) = \sum_{k=1}^{15} k \cdot { }^{15} C_{k} \cdot x^{k-1}\)

Therefore,

\(15(1 + x)^{14} = \sum_{k=1}^{15} k \cdot { }^{15} C_{k} \cdot x^{k-1}\)

Setting \(x = -1\):

\(\Rightarrow 15 \cdot (1 - 1)^{14} = 0 = \sum_{k=1}^{15} (-1)^{k} \cdot k \cdot { }^{15} C_{k}\)

2. Evaluate the Second Part:

Using the combinatorial identity for the sum of binomial coefficients at odd indices:

\((1+x)^{n} = \sum_{k=0}^{n} { }^{n} C_{k} \cdot x^{k}\)

Set \(x = 1\) and \(x = -1\):

\(\Rightarrow (1+1)^{14} = \sum_{k=0}^{14} { }^{14} C_{k}\) and \((1-1)^{14} = \sum_{k=0}^{14}(-1)^{k} \cdot { }^{14} C_{k}\)

Thus, \(\Rightarrow 2^{14} = { }^{14} C_{0} + { }^{14} C_{2} + { }^{14} C_{4} + \ldots\) and \(0 = { }^{14} C_{0} - { }^{14} C_{1} + { }^{14} C_{2} - \ldots\).

Adding the two equations, the sum of coefficients at odd positions:

\(2^{13} = ({ }^{14} C_{1} + { }^{14} C_{3} + { }^{14} C_{5} + \ldots)\)

Final Calculation:

Combine both parts:

\(0 + (2^{13}) = 2^{13}\)

The original expression also states \(+ { }^{14} C _{13}\) which is included in \(2^{13}\).

Hence, the final answer is:

Correct Answer: \(2^{13} - 14\)