Question

$1 + {}^{100}C_1 + {}^{100}C_2 + \dots + {}^{100}C_{99} + 1 =$

• A. $2^{99}$
• B. $2^{101}$
• C. $2^{98}$
• D. $2^{100}$
• E. $100^2$

Hint:

Remember ${}^nC_0 = 1$ and ${}^nC_n = 1$ to recognize complete binomial sums.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question relates to the properties of binomial coefficients and the binomial theorem. The sum of all binomial coefficients for a given \( n \) has a specific formula.
Step 2: Key Formula or Approach:
The binomial theorem states that for any integer \( n \geq 0 \):
\[ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n}y^n \] A useful identity is derived by setting \( x=1 \) and \( y=1 \):
\[ (1+1)^n = 2^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} \] We also need the identities for the first and last binomial coefficients: \( \binom{n}{0} = 1 \) and \( \binom{n}{n} = 1 \).
Step 3: Detailed Explanation:
The given expression is:
\[ S = 1 + ^{100}C_1 + ^{100}C_2 + \dots + ^{100}C_{99} + 1 \] Let's rewrite the expression using the properties of binomial coefficients. We know that for \( n=100 \):
\( ^{100}C_0 = 1 \)
\( ^{100}C_{100} = 1 \)
We can substitute these into the given expression. The first '1' can be written as \( ^{100}C_0 \) and the last '1' can be written as \( ^{100}C_{100} \).
So, the expression becomes:
\[ S = ^{100}C_0 + ^{100}C_1 + ^{100}C_2 + \dots + ^{100}C_{99} + ^{100}C_{100} \] This is the sum of all binomial coefficients for \( n=100 \).
Using the identity from Step 2:
\[ \sum_{k=0}^{100} \binom{100}{k} = 2^{100} \] Therefore, the value of the expression is \( 2^{100} \).
Step 4: Final Answer:
The sum is equal to \( 2^{100} \).