Solve in english

The equation for $L_1$ is $\frac{x+2}{4} + \frac{y-1}{m} = \frac{3}{4}$.
To eliminate denominators, multiply the entire equation by $4m$:

Rearrange into the form $Ax + By = C$:

The equation for $L_2$ is given in vector form: $\mathbf{r} = 2\mathbf{i} + 3\mathbf{j} + t(\mathbf{i} + m\mathbf{j})$.
This means $x = 2 + t$ and $y = 3 + mt$.
From the first equation, $t = x-2$. Substitute this into the second equation:

Rearrange into the form $Ax + By = C$:

For two lines to intersect, they must not be parallel and distinct.
Lines are parallel if their slopes are equal. From $mx + 4y = m+4$, the slope is $-\frac{m}{4}$. From $mx - y = 2m-3$, the slope is $m$.
For the lines to be parallel, $-\frac{m}{4} = m$.

If $m=0$, the lines are parallel. Let's check if they are distinct or identical when $m=0$:
For $m=0$, equation $(*)$ becomes:

For $m=0$, equation $(**)$ becomes:

Since $y=1$ and $y=3$ are two different horizontal lines, they are parallel and distinct. Therefore, they do not intersect when $m=0$.

For any other value of $m$ (i.e., $m \ne 0$), the slopes will not be equal, meaning the lines are not parallel. Non-parallel lines in a 2D plane always intersect at a unique point.
Thus, the lines $L_1$ and $L_2$ intersect for all values of $m$ except $m=0$.

The value of $m$ for which $L_1$ and $L_2$ intersect is $\boxed{m \ne 0}$.

Let $P(n)$ be the statement $\sum_{r=1}^{n} \frac{r}{2^r} = 2 - \frac{n+2}{2^n}$.
For $n=1$:
Left Hand Side (LHS): $\sum_{r=1}^{1} \frac{r}{2^r} = \frac{1}{2^1} = \frac{1}{2}$.
Right Hand Side (RHS): $2 - \frac{1+2}{2^1} = 2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{1}{2}$.
Since LHS = RHS, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k \ge 1$.
That is, assume $\sum_{r=1}^{k} \frac{r}{2^r} = 2 - \frac{k+2}{2^k}$.

We need to prove that $P(k+1)$ is true.
That is, we need to show $\sum_{r=1}^{k+1} \frac{r}{2^r} = 2 - \frac{(k+1)+2}{2^{k+1}} = 2 - \frac{k+3}{2^{k+1}}$.

Start with the LHS of $P(k+1)$:

Using the inductive hypothesis for the sum $\sum_{r=1}^{k} \frac{r}{2^r}$:

To combine the fractional terms, find a common denominator, which is $2^{k+1}$:

Combine the fractions:

This matches the RHS of $P(k+1)$.
Therefore, $P(k+1)$ is true.

By the principle of mathematical induction, the statement $\sum_{r=1}^{n} \frac{r}{2^r} = 2 - \frac{n+2}{2^n}$ is true for all integers $n \ge 1$.