Question

\(2+{}^{15}C_{1}+{}^{15}C_{2}+\dots+{}^{15}C_{14}=\)

• A. \(2^{14}\)
• B. \(2^{15}\)
• C. \(2^{16}\)
• D. \(2^{10}\)
• E. \(2^{18}\)

Hint:

Whenever you see a long sum of binomial coefficients, try to convert it into the complete form \(\sum_{r=0}^{n} {}^{n}C_r\). Then directly use \((1+1)^n=2^n\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the sum of binomial coefficients. We need to use the property of the binomial expansion of \((1+x)^n\).
Step 2: Key Formula or Approach:
The Binomial Theorem states that \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\).
A key identity derived from this by setting a=1 and b=1 is the sum of all binomial coefficients for a given n:
\[ \sum_{r=0}^{n} {^{n}C_r} = {^{n}C_0} + {^{n}C_1} + {^{n}C_2} + \dots + {^{n}C_n} = 2^n \] Step 3: Detailed Explanation:
Let's analyze the given expression: \(E = 2 + {^{15}C_1} + {^{15}C_2} + \dots + {^{15}C_{14}}\).
The sum part of the expression is \(S = {^{15}C_1} + {^{15}C_2} + \dots + {^{15}C_{14}}\).
From the key formula, for n=15, we have:
\[ {^{15}C_0} + {^{15}C_1} + {^{15}C_2} + \dots + {^{15}C_{14}} + {^{15}C_{15}} = 2^{15} \] The sum \(S\) is missing the first term (\(^{15}C_0\)) and the last term (\(^{15}C_{15}\)) from the full expansion.
We know that:
\[ {^{n}C_0} = 1 \quad \text{and} \quad {^{n}C_n} = 1 \] So, \(^{15}C_0 = 1\) and \(^{15}C_{15} = 1\).
We can express \(S\) in terms of the full sum:
\[ S = ({^{15}C_0} + {^{15}C_1} + \dots + {^{15}C_{14}} + {^{15}C_{15}}) - {^{15}C_0} - {^{15}C_{15}} \] \[ S = 2^{15} - 1 - 1 = 2^{15} - 2 \] Now substitute this back into the original expression for \(E\):
\[ E = 2 + S = 2 + (2^{15} - 2) \] \[ E = 2^{15} \] Step 4: Final Answer:
The value of the expression is \(2^{15}\). Therefore, option (B) is the correct answer.