COMEDK · Maths · 7. Binomial Theorem
If \(c_{0}, c_{1}, c_{2}, \ldots, c_{n}\) are binomial coefficients of order \(n\), then the value of \(\frac{c_{1}}{2}+\frac{c_{2}}{4}+\frac{c_{3}}{6}+\ldots\)
- A \(\frac{2^{n}+1}{n-1}\)
- B \(\frac{2^{n}-1}{n+1}\)
- C \(\frac{2^{n}+1}{n-1}\)
- D \(\frac{2^{n}}{n+1}\)
Answer & Solution
Correct Answer
(B) \(\frac{2^{n}-1}{n+1}\)
Step-by-step Solution
Detailed explanation
We have, \(c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots=\frac{1}{2}\left\{(1+x)^{n}-(1-x)^{n}\right\}\) \(\Rightarrow \int_{0}^{1}\left(c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots\right) d x\) \(=\frac{1}{2} \int_{0}^{1}(1+x)^{n}-(1-x)^{n} d x\)…
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