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
- $2^{16}-1$
- $2^{13}-14$
- $2^{14}$
- $2^{13}-13$
The Correct Option is B
Solution and Explanation
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:
- \(-{ }^{15} C_{1} + 2 \cdot { }^{15} C_{2} - 3 \cdot { }^{15} C_{3} + \ldots - 15 \cdot { }^{15} C_{15}\)
- \({ }^{14} C_{1} + { }^{14} C_{3} + { }^{14} C_{5} + \ldots + { }^{14} C_{11}\)
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: \(2^{13} - 14\)
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Concepts Used:
Combinations
The method of forming subsets by selecting data from a larger set in a way that the selection order does not matter is called the combination.
- It means the combination of about ‘n’ things taken ‘k’ at a time without any repetition.
- The combination is used for a group of data where the order of data does not matter.
- For example, Imagine you go to a restaurant and order some soup.
- Five toppings can complement the soup, namely:
- croutons,
- orange zest,
- grated cheese,
- chopped herbs,
- fried noodles.
But you are only allowed to pick three.
- There can be several ways in which you can enhance your soup with savory.
- The selection of three toppings (subset) from the five toppings (larger set) is called a combination.
Use of Combinations:
It is used for a group of data (where the order of data doesn’t matter).