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: Interpret the given expression carefully.
The expression is:
\[ 2+{}^{15}C_{1}+{}^{15}C_{2}+\dots+{}^{15}C_{14} \] Now note that the first number \(2\) can be written as:
\[ 2={} ^{15}C_0+{}^{15}C_{15} \] because \[ {}^{15}C_0=1 \quad \text{and} \quad {}^{15}C_{15}=1 \] So, \[ {}^{15}C_0+{}^{15}C_{15}=1+1=2 \]

Step 2: Rewrite the whole sum using binomial coefficients.
Substituting this into the given expression, we get:
\[ 2+{}^{15}C_{1}+{}^{15}C_{2}+\dots+{}^{15}C_{14} = {}^{15}C_0+{}^{15}C_1+{}^{15}C_2+\dots+{}^{15}C_{14}+{}^{15}C_{15} \]

Step 3: Recognize the complete binomial expansion sum.
Now the right-hand side is the sum of all binomial coefficients of order \(15\):
\[ \sum_{r=0}^{15} {}^{15}C_r \]

Step 4: Recall the standard identity.
From the binomial theorem, we know:
\[ (1+1)^{15}=\sum_{r=0}^{15} {}^{15}C_r \] Therefore,
\[ \sum_{r=0}^{15} {}^{15}C_r=2^{15} \]

Step 5: Apply the identity to the given sum.
Hence,
\[ {}^{15}C_0+{}^{15}C_1+{}^{15}C_2+\dots+{}^{15}C_{14}+{}^{15}C_{15}=2^{15} \] So the given expression is equal to:
\[ 2^{15} \]

Step 6: Verify the logic once more.
The only trick in this question is to notice that the initial \(2\) must be split as:
\[ 2=1+1={} ^{15}C_0+{}^{15}C_{15} \] After that, the whole expression becomes the complete sum of coefficients in \((1+1)^{15}\).

Step 7: Match with the options.
Thus, the required value is:
\[ \boxed{2^{15}} \] which matches option \((2)\).