Question:
If \( C_{o},C_{1},C_{2},...,C_{n} \) represent the coefficients in the binomial expansion of \( (1+x)^{n} \), then \( C_{o}+\frac{c_{2}}{3}+\frac{c_{4}}{5}+\cdot\cdot\cdot+\frac{c_{16}}{17}= \)
If \( C_{o},C_{1},C_{2},...,C_{n} \) represent the coefficients in the binomial expansion of \( (1+x)^{n} \), then \( C_{o}+\frac{c_{2}}{3}+\frac{c_{4}}{5}+\cdot\cdot\cdot+\frac{c_{16}}{17}= \)
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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 \( n + 1 \).
Updated On: Jun 8, 2026
- \( \frac{2^{14}}{17} \)
- \( \frac{2^{15}}{17} \)
- \( \frac{2^{16}}{17} \)
- \( \frac{2^{17}}{17} \)
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The Correct Option is C
Solution and Explanation
Concept:
The given series consists of alternate binomial coefficients divided by successive odd integers. Let us look at the standard integral property of binomial series:
\[
\int (1+x)^n dx = \frac{(1+x)^{n+1}}{n+1}
\]
By evaluating this integration between specific symmetric bounds or separating odd and even components, we find:
\[
C_0 + \frac{C_2}{3} + \frac{C_4}{5} + \dots = \frac{2^n}{n+1}
\]
Step 1: Identifying the value of \( n \) from the last term.
The last term of the series is given as \( \frac{C_{16}}{17} \). The general format of the denominator is \( 2k + 1 \) when the coefficient index is \( 2k \). For the final term, \( 2k = 16 \implies 2k+1 = 17 \). This implies that the upper index of the binomial expansion matches the final boundary value: \[ n = 16 \]
Step 2: Applying the standard summation formula.
Substitute \( n = 16 \) into the alternate coefficient formula: \[ \text{Sum} = \frac{2^n}{n+1} = \frac{2^{16}}{16+1} = \frac{2^{16}}{17} \] This perfectly matches option (C).
Step 1: Identifying the value of \( n \) from the last term.
The last term of the series is given as \( \frac{C_{16}}{17} \). The general format of the denominator is \( 2k + 1 \) when the coefficient index is \( 2k \). For the final term, \( 2k = 16 \implies 2k+1 = 17 \). This implies that the upper index of the binomial expansion matches the final boundary value: \[ n = 16 \]
Step 2: Applying the standard summation formula.
Substitute \( n = 16 \) into the alternate coefficient formula: \[ \text{Sum} = \frac{2^n}{n+1} = \frac{2^{16}}{16+1} = \frac{2^{16}}{17} \] This perfectly matches option (C).
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