Question

Let f(x) = \sum_{n=1}^\infty \left( \frac{2^{2n-1}}{1+x^{2n}} \right) \left( \frac{x^{2^{n-1}}}{1} \right) for x \in (0, 1), then f'\left(\frac{1}{4}\right) is ____.

Answer 1

544/225

Answer 2

Question 9: Solution

Let's assume the question has a typo and the intended function is $f(x) = \sum_{n=1}^\infty \frac{2^n x^{2^n-1}}{1+x^{2^n}}$. This is a common type of series in competitive exams that simplifies.

We know the identity derived from the product $\prod_{k=0}^\infty (1+x^{2^k}) = \frac{1}{1-x}$. Taking the natural logarithm of both sides: $\sum_{k=0}^\infty \ln(1+x^{2^k}) = \ln\left(\frac{1}{1-x}\right) = -\ln(1-x)$. Differentiating both sides with respect to $x$: $\sum_{k=0}^\infty \frac{1}{1+x^{2^k}} \cdot (2^k x^{2^k-1}) = \frac{1}{1-x}$. Let's split the sum: $\frac{2^0 x^{2^0-1}}{1+x^{2^0}} + \sum_{k=1}^\infty \frac{2^k x^{2^k-1}}{1+x^{2^k}} = \frac{1}{1-x}$. $\frac{1}{1+x} + \sum_{k=1}^\infty \frac{2^k x^{2^k-1}}{1+x^{2^k}} = \frac{1}{1-x}$. Therefore, $f(x) = \sum_{k=1}^\infty \frac{2^k x^{2^k-1}}{1+x^{2^k}} = \frac{1}{1-x} - \frac{1}{1+x} = \frac{(1+x)-(1-x)}{(1-x)(1+x)} = \frac{2x}{1-x^2}$. Now we need to find $f'(x)$. $f'(x) = \frac{d}{dx} \left( \frac{2x}{1-x^2} \right)$. Using the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$. Here, $u=2x$ and $v=1-x^2$. So $u'=2$ and $v'=-2x$. $f'(x) = \frac{2(1-x^2) - 2x(-2x)}{(1-x^2)^2} = \frac{2-2x^2+4x^2}{(1-x^2)^2} = \frac{2+2x^2}{(1-x^2)^2} = \frac{2(1+x^2)}{(1-x^2)^2}$. Finally, we need to evaluate $f'\left(\frac{1}{4}\right)$: $f'\left(\frac{1}{4}\right) = \frac{2\left(1+\left(\frac{1}{4}\right)^2\right)}{\left(1-\left(\frac{1}{4}\right)^2\right)^2} = \frac{2\left(1+\frac{1}{16}\right)}{\left(1-\frac{1}{16}\right)^2} = \frac{2\left(\frac{16+1}{16}\right)}{\left(\frac{16-1}{16}\right)^2} = \frac{2\left(\frac{17}{16}\right)}{\left(\frac{15}{16}\right)^2}$. $f'\left(\frac{1}{4}\right) = \frac{\frac{17}{8}}{\frac{225}{256}} = \frac{17}{8} \cdot \frac{256}{225} = \frac{17 \cdot 32}{225} = \frac{544}{225}$.

Explanation of the solution for Q9 (minimal):

1. Assume the question intended $f(x) = \sum_{n=1}^\infty \frac{2^n x^{2^n-1}}{1+x^{2^n}}$ due to common problem patterns and expected complexity.
2. Use the identity $\sum_{k=0}^\infty \frac{2^k x^{2^k-1}}{1+x^{2^k}} = \frac{1}{1-x}$ (derived from differentiating $\ln(\prod (1+x^{2^k})) = \ln(\frac{1}{1-x})$).
3. Extract the sum from $n=1$: $f(x) = \frac{1}{1-x} - \frac{1}{1+x} = \frac{2x}{1-x^2}$.
4. Differentiate $f(x)$: $f'(x) = \frac{d}{dx}\left(\frac{2x}{1-x^2}\right) = \frac{2(1-x^2) - 2x(-2x)}{(1-x^2)^2} = \frac{2+2x^2}{(1-x^2)^2}$.
5. Substitute $x=\frac{1}{4}$: $f'\left(\frac{1}{4}\right) = \frac{2(1+(1/4)^2)}{(1-(1/4)^2)^2} = \frac{2(17/16)}{(15/16)^2} = \frac{17/8}{225/256} = \frac{17 \cdot 32}{225} = \frac{544}{225}$.