Find the Taylor series expansion of 1/√(x-8)² up to the term in (x-8)².

Let $\frac{5}{(5r-1)(5r+4)} = \frac{A}{5r-1} + \frac{B}{5r+4}$.
Multiplying by $(5r-1)(5r+4)$, we get $5 = A(5r+4) + B(5r-1)$.
Set $5r-1 = 0 \implies r = \frac{1}{5}$:
$5 = A(5(\frac{1}{5})+4) \implies 5 = A(1+4) \implies 5 = 5A \implies A = 1$.
Set $5r+4 = 0 \implies r = -\frac{4}{5}$:
$5 = B(5(-\frac{4}{5})-1) \implies 5 = B(-4-1) \implies 5 = -5B \implies B = -1$.
So, the general term is $\frac{1}{5r-1} - \frac{1}{5r+4}$.

For $r=1$: $\frac{1}{5(1)-1} - \frac{1}{5(1)+4} = \frac{1}{4} - \frac{1}{9}$
For $r=2$: $\frac{1}{5(2)-1} - \frac{1}{5(2)+4} = \frac{1}{9} - \frac{1}{14}$
For $r=3$: $\frac{1}{5(3)-1} - \frac{1}{5(3)+4} = \frac{1}{14} - \frac{1}{19}$
...
For $r=n$: $\frac{1}{5n-1} - \frac{1}{5n+4}$

When summed, the intermediate terms cancel out:

Thus, it is shown that $\boxed{\sum_{r=1}^{n} \frac{5}{(5r-1)(5r+4)} = \frac{5n}{4(5n+4)}}$.

9b)
To find the Taylor series expansion of $f(x) = \frac{1}{\sqrt{x}}$ up to the term in $(x-8)^2$, we need to evaluate $f(x)$, $f'(x)$, and $f''(x)$ at $x=8$.
The Taylor series formula around $x=a$ is $f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots$. Here $a=8$.

The Taylor series expansion is $\boxed{\frac{\sqrt{2}}{4} - \frac{\sqrt{2}}{64}(x-8) + \frac{3\sqrt{2}}{2048}(x-8)^2}$.

9c)
To prove by induction that $3^{4n} + 2^{4n+2}$ is divisible by 5 for all positive integers $n$.

For $n=1$, we have $3^{4(1)} + 2^{4(1)+2} = 3^4 + 2^6 = 81 + 64 = 145$.
Since $145 = 5 \times 29$, 145 is divisible by 5.
Thus, the statement is true for $n=1$.

Assume the statement is true for some positive integer $k$.
That is, assume $3^{4k} + 2^{4k+2}$ is divisible by 5.
This means $3^{4k} + 2^{4k+2} = 5m$ for some integer $m$.
From this, we can write $3^{4k} = 5m - 2^{4k+2}$.

We need to show that $3^{4(k+1)} + 2^{4(k+1)+2}$ is divisible by 5.

Substitute $3^{4k} = 5m - 2^{4k+2}$ from the inductive hypothesis:

Since $65 = 5 \times 13$, we can factor out 5:

Since $(81m - 13 \cdot 2^{4k+2})$ is an integer, $3^{4(k+1)} + 2^{4(k+1)+2}$ is divisible by 5.

By the principle of mathematical induction, $3^{4n} + 2^{4n+2}$ is divisible by 5 for all positive integers $n$.