Use mathematical induction to prove that 2+6+10+...+(4n+2) = 2n(n+1) for all positive integers n.

The statement to prove is $2+6+10+\dots+(4n+2) = 2n(n+1)$ for all positive integers $n$.

First, let's analyze the sum. The terms $2, 6, 10, \dots$ form an arithmetic progression with first term $a_1=2$ and common difference $d=4$. The $k$-th term of this sequence is $a_k = a_1 + (k-1)d = 2 + (k-1)4 = 4k-2$.
The last term given in the sum is $(4n+2)$. To find out which term this is, we set $a_k = 4n+2$:
$4k-2 = 4n+2$
$4k = 4n+4$
$k = n+1$.
This means the sum is actually the sum of the first $(n+1)$ terms of the sequence $a_k = 4k-2$.
The sum of the first $m$ terms of an arithmetic progression is $S_m = \frac{m}{2}(a_1 + a_m)$.
Here, $m=n+1$, $a_1=2$, and $a_m = a_{n+1} = 4(n+1)-2 = 4n+4-2 = 4n+2$.
So, the sum is $S_{n+1} = \frac{n+1}{2}(2 + (4n+2)) = \frac{n+1}{2}(4n+4) = \frac{n+1}{2} \cdot 4(n+1) = 2(n+1)^2$.

The problem asks to prove $2(n+1)^2 = 2n(n+1)$.
Let's check this equality:
$2(n+1)^2 = 2(n^2+2n+1) = 2n^2+4n+2$.
$2n(n+1) = 2n^2+2n$.
Since $2n^2+4n+2 \neq 2n^2+2n$ (they are only equal if $2n+2=0$, which means $n=-1$, not a positive integer), the statement as written is incorrect.

It is highly probable that there is a typo in the question, and the intended statement to prove is $2+6+10+\dots+(4n+2) = 2(n+1)^2$. I will proceed with proving this corrected statement.

Statement to prove (corrected): $P(n): 2+6+10+\dots+(4n+2) = 2(n+1)^2$ for all positive integers $n$.

The Taylor expansion of a function $f(x)$ about $x=a$ is given by:

Here, $f(x) = \ln x$ and $a=1$.