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If $c_{0}, c_{1}, c_{2}, \ldots, c_{n}$ are binomial coefficients of order $n$, then the value of $\frac{c_{1}}{2}+\frac{c_{2}}{4}+\frac{c_{3}}{6}+\ldots$
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Verified Answer
The correct answer is:
$\frac{2^{n}-1}{n+1}$
We have,
$c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots=\frac{1}{2}\left\{(1+x)^{n}-(1-x)^{n}\right\}$
$\Rightarrow \int_{0}^{1}\left(c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots\right) d x$
$=\frac{1}{2} \int_{0}^{1}(1+x)^{n}-(1-x)^{n} d x$
$\Rightarrow \frac{c_{1}}{2}+\frac{c_{3}}{4}+\frac{c_{5}}{6}+\ldots=\frac{2^{n}-1}{n+1}$
$\log _{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{4.2}+\frac{1}{4}\left(\frac{1}{2}\right)^{2}+\ldots \infty\right)$
$c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots=\frac{1}{2}\left\{(1+x)^{n}-(1-x)^{n}\right\}$
$\Rightarrow \int_{0}^{1}\left(c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots\right) d x$
$=\frac{1}{2} \int_{0}^{1}(1+x)^{n}-(1-x)^{n} d x$
$\Rightarrow \frac{c_{1}}{2}+\frac{c_{3}}{4}+\frac{c_{5}}{6}+\ldots=\frac{2^{n}-1}{n+1}$
$\log _{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{4.2}+\frac{1}{4}\left(\frac{1}{2}\right)^{2}+\ldots \infty\right)$
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