Question

If \( C_{o},C_{1},C_{2},...,C_{n} \) represent the coefficients in the binomial expansion of \( (1+x)^{n} \), then \( C_{o}+\frac{c_{2}}{3}+\frac{c_{4}}{5}+\cdot\cdot\cdot+\frac{c_{16}}{17}= \)

• A. \( \frac{2^{14}}{17} \)
• B. \( \frac{2^{15}}{17} \)
• C. \( \frac{2^{16}}{17} \)
• D. \( \frac{2^{17}}{17} \)

Hint:

Whenever binomial coefficients are divided by numbers in an arithmetic progression (\( 1, 3, 5, \dots \)), it is a direct indicator of integration. The final denominator always becomes \( n + 1 \).

Answer 1

The Correct Option is C

Step 1: Look at the pattern.
The series is $C_0 + \dfrac{C_2}{3} + \dfrac{C_4}{5} + \cdots$, using even-numbered coefficients over odd numbers.
Step 2: Recall the integral trick.
Integrating $(1+x)^n$ and using symmetric limits separates the even and odd coefficients, giving the known result \[ C_0 + \frac{C_2}{3} + \frac{C_4}{5} + \cdots = \frac{2^n}{n+1} \]
Step 3: Find n from the last term.
The last term is $\dfrac{C_{16}}{17}$. The denominator $17$ comes from index $16$ being the top, so $n = 16$.
Step 4: Plug into the formula.
\[ \text{Sum} = \frac{2^{16}}{16+1} = \frac{2^{16}}{17} \]
Step 5: Confirm the denominator.
The $n+1 = 17$ matches the last denominator, so the setup is consistent.
Step 6: State the answer.
The sum equals \[ \boxed{\frac{2^{16}}{17}} \]