一个级数的前 $n$ 项由 $\log_a 3 + \log_a 27 + \log_a 243 + \ldots + \log_a 3^{2^{n-1}}$ 给出，其中 $a$ 是一个正数常数。 证明该级数是等差级数。 已知该级数前 30 项的和为 300，求 $a$ 的值。 一个等比级数的首项为 $c$，公比为 $r$，其中 $c$ 和 $r$ 均不为零。一个等差级数的首项为 $b$，公差为 $d$，其中 $b$ 和 $d$ 均不为零。已知该等差级数的第 5 项、第 8 项和第 10 项分别等于该等比级数的第 2 项、第 3 项和第 4 项。证明 $r$ 满足方程 $3r^2 - 5r + 2 = 0$，并由此用 $c$ 表示其无穷项和。

1. Rewrite the terms using the power law of logarithms

The terms are $\log_a 3 = \log_a 3^1$, $\log_a 27 = \log_a 3^3$, $\log_a 243 = \log_a 3^5$, with the $n$th term $\log_a 3^{2n-1}$.

Using $\log_a x^k = k\log_a x$:
$T_n = \log_a 3^{2n-1} = (2n-1)\log_a 3.$
So the terms are $\log_a 3,\; 3\log_a 3,\; 5\log_a 3,\; \ldots$

2. Show the common difference is constant

Consider the difference between consecutive terms:
$T_n - T_{n-1} = (2n-1)\log_a 3 - (2(n-1)-1)\log_a 3 = \big[(2n-1)-(2n-3)\big]\log_a 3 = 2\log_a 3.$
This difference is the same constant for all $n$, so the series is an arithmetic series (with common difference $2\log_a 3$).

1. Identify the first term and common difference of the arithmetic series.

From the given series, the first term is $u_1 = \log_a 3$.
The second term is $u_2 = \log_a 27 = \log_a 3^3 = 3 \log_a 3$.
The common difference is $d = u_2 - u_1 = 3 \log_a 3 - \log_a 3 = 2 \log_a 3$.

2. Form an equation using the sum of the first 30 terms and solve for $a$.

Using the sum formula $S_n = \frac{n}{2}[2u_1 + (n-1)d]$ for $n=30$:
$S_{30} = \frac{30}{2}[2(\log_a 3) + 29(2 \log_a 3)]$
$S_{30} = 15[2 \log_a 3 + 58 \log_a 3] = 15(60 \log_a 3) = 900 \log_a 3$
Equating this to the given sum of 300:
$900 \log_a 3 = 300$
$\log_a 3 = \frac{1}{3}$
Converting from logarithmic form to index form:
$a^{\frac{1}{3}} = 3$
$a = 3^3 = 27$

1. Express the given terms as equations.

Let the 5th, 8th, and 10th terms of the arithmetic series be equal to the 2nd, 3rd, and 4th terms of the geometric series respectively:
$b + 4d = cr$ --- (1)
$b + 7d = cr^2$ --- (2)
$b + 9d = cr^3$ --- (3)

2. Eliminate $b$ to find expressions relating $d$, $c$, and $r$.

Subtract (1) from (2):
$(b + 7d) - (b + 4d) = cr^2 - cr$
$3d = c(r^2 - r)$ --- (4)

Subtract (2) from (3):
$(b + 9d) - (b + 7d) = cr^3 - cr^2$
$2d = c(r^3 - r^2)$ --- (5)

3. Eliminate $d$ to deduce the given quadratic equation in $r$.

From (4), $d = \frac{c(r^2 - r)}{3}$
From (5), $d = \frac{c(r^3 - r^2)}{2}$

Equating the expressions for $d$:
$\frac{c(r^2 - r)}{3} = \frac{c(r^3 - r^2)}{2}$
$2c(r^2 - r) = 3c(r^3 - r^2)$

Since $c \neq 0$, we can divide both sides by $c$:
$2r^2 - 2r = 3r^3 - 3r^2$
$3r^3 - 5r^2 + 2r = 0$

Since $r \neq 0$, we can divide the equation by $r$:
$3r^2 - 5r + 2 = 0$ (Shown)

4. Solve the quadratic equation and reject the invalid value for $r$.

Factorising the quadratic equation:
$(3r - 2)(r - 1) = 0$
$r = \frac{2}{3} \text{ or } r = 1$

If $r = 1$, substituting into equation (4) gives $3d = c(1^2 - 1) = 0$, which means $d = 0$.
However, it is given that $d \neq 0$, so we reject $r = 1$.
Therefore, $r = \frac{2}{3}$.

5. Calculate the sum to infinity of the geometric series.

The sum to infinity $S_\infty$ of the geometric series is given by $\frac{\text{first term}}{1 - r}$:
$S_\infty = \frac{c}{1 - r}$
$S_\infty = \frac{c}{1 - \frac{2}{3}}$
$S_\infty = \frac{c}{\frac{1}{3}} = 3c$