Sequences and series (1)

We have

$\begin{aligned} f(x) &= 7 + 2x \ln(25) - 5^{x-1} - 5^{2-x} & \small {\color{#3D99F6} \text{Differentiating w.r.t. } x \text{ and equating to zero}} \\ \implies f'(x) &= 2 \ln(25) - 5^{x-1} \ln(5) + 5^{2-x} \ln(5) = 0 \\ &\implies 4 \ln(5) - 5^{x-1} \ln(5) + 5^{2-x} \ln(5) = 0 \\ &\implies 4 - 5^{x-1} + 5^{2-x} =0 \\ &\implies 5^{x-1} - 5^{2-x} = 4 \\ &\implies x = 2 \end{aligned}$

Checking the second derivative, we find that

$\begin{aligned} & f''(x) = - 5^{x-1} \ln^2 (5) - 5^{2-x} \ln^2 (5) \\ \implies & f''(2) = - 6 \ln^2 (5) < 0 \end{aligned}$

Thus $x=2$ is the point of local maxima for $f(x)$ and we have

$\color{#D61F06} a = 2$

Now

$\begin{aligned} \displaystyle \lim_{x \to 0} \int_0^x \dfrac{t^2 \,dt}{x^2 \tan (\pi+x)} &= \lim_{x \to 0} \dfrac{1}{x^2 \tan (\pi+x)} \int_0^x t^2 \,dt \\ &= \displaystyle \lim_{x \to 0} \dfrac{1}{x^2 \tan (\pi+x)} \cdot \dfrac{x^3}{3} \\ &= \displaystyle \lim_{x \to 0} \dfrac{x}{3 \tan (\pi +x)} & \small {\color{#3D99F6} \dfrac 00 \text{ case, L'Hopital's Rule applies}} \\ &= \displaystyle \lim_{x \to 0} \dfrac{1}{3 \sec^2 (\pi +x)} \\ &= \dfrac 13 \end{aligned}$

Hence

$\color{#D61F06} r = \dfrac 13$

Note that $|r| = \dfrac 13 < 1$ and therefore the sum our infinite geometric progression converges and is equal to

$S = a + ar + ar^2 + ar^3 + \cdots = \dfrac{a}{1-r} = \dfrac{2}{1 - \frac 13} = \boxed{3}$