1. Geometric progression?

Find the sum of the terms of the G.P., $a + ar + ar^2 + ...\ \infty$ where $a$ is the value of $x$ for which the function $7 + 2x\log_{e}25 - 5^{x - 1} - 5^{2 - x}$ has the greatest value and $r$ is the limit $\lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt$

2. Originally Posted by fardeen_gen
Find the sum of the terms of the G.P., $a + ar + ar^2 + ...\ \infty$ where $a$ is the value of $x$ for which the function $7 + 2x\log_{e}25 - 5^{x - 1} - 5^{2 - x}$ has the greatest value and $r$ is the limit $\lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt$
First

$r = \lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt = \lim_{x \to 0} \frac{\frac{1}{3}x^3}{x^2\tan(\pi + x)} = \frac{1}{3} \lim_{x \to 0} \frac{x}{\tan(\pi + x)} = \frac{1}{3}$ by L'Hopitals rule.

If we let $y = 7 + 4x\ln 5 - 5^{x - 1} - 5^{2 - x}$ then using calculus we find the max is located at $x = 2$ giving a maximum of

$
a = 8 \ln 5 + 1
$

Since $r <$1, then the sum of the series is $S_\infty = \frac{a}{1-r} = \frac{3}{2} (8 \ln 5 + 1)$

3. Answer is 3 according to the problem book. Have they taken any assumptions?

4. Originally Posted by danny arrigo
First

$r = \lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt = \lim_{x \to 0} \frac{\frac{1}{3}x^3}{x^2\tan(\pi + x)} = \frac{1}{3} \lim_{x \to 0} \frac{x}{\tan(\pi + x)} = \frac{1}{3}$ by L'Hopitals rule.

If we let $y = 7 + 4x\ln 5 - 5^{x - 1} - 5^{2 - x}$ then using calculus we find the max is located at $x = 2$ giving a maximum of

$
a = 8 \ln 5 + 1
$

Since $r <$1, then the sum of the series is $S_\infty = \frac{a}{1-r} = \frac{3}{2} (8 \ln 5 + 1)$
I miss read the question, $a = 2$, the $x$ value of the maximum. I originally used the actual maximum value. So

$
S = \frac{a}{1-r} = \frac{2}{1 - \frac{1}{3}} = 3
$