MaxWong

We notice that the n-th term is $\dfrac{n}{(n + 1)!}$

If we did some sort of arithmetic trick on it:

$\dfrac{n}{(n + 1)!} = \dfrac{(n + 1) - 1}{(n + 1)!} = \dfrac{(n + 1)}{(n + 1)(n!)} - \dfrac{1}{(n + 1)!} = \dfrac1{n!} - \dfrac1{(n + 1)!}$

That means, the sum is $\displaystyle\sum_{n = 1}^\infty \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right)$.

To evaluate general infinite sums, we consider the partial sum first, then take the limit. 

So first, we consider $\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right)$.

Listing the terms, 

$\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right) = \left(\dfrac1{1!} - \dfrac1{2!}\right) + \left(\dfrac1{2!} - \dfrac1{3!}\right) + \left(\dfrac1{3!} - \dfrac1{4!}\right) + \cdots + \left(\dfrac1{N!} - \dfrac1{(N+1)!}\right)$

As you may have noticed, the terms in the middle eliminates each other nicely. The only terms left are the first, and the last.

$\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right) = \dfrac1{1!} - \dfrac1{(N + 1)!} = 1 - \dfrac1{(N + 1)!}$

Next, we will take $\displaystyle\lim_{N\to \infty}$ on both sides.

(If you don't know limits, don't panic. The above notation just means "N is a really large number".)

$\displaystyle\lim_{N\to \infty}\displaystyle\sum_{n = 1}^N \left(\dfrac1{n!} - \dfrac1{(n + 1)!}\right)= \displaystyle\lim_{N\to \infty}\left(1 - \dfrac1{(N + 1)!}\right)\\ \displaystyle\sum_{n=1}^\infty \dfrac{n}{(n + 1)!} = 1 - \dfrac1{\text{Really big number}} = 1$

The required sum is 1.

Observation: The grid in the top left corner of a rectangle of given length and width, defines the rectangle.

(Alternatively, this means that given the size of the rectangle, and given the top left grid, you can only specify one rectangle. None of the other possible rectangles have the same size, or have the same top left grid.)

Because of this observation, if we change the top left corner of the 3×2 rectangle, we immediately know that it is different from the previous rectangle.

But 2×3 rectangles should also count. So we have to consider them too.

For 3×2 and 2×3 rectangles, each of the types have 12 possible top left corners. Can you count them? 

Answer = 12 + 12 = 24

Notice that the left hand side is a G.P. with first term = 1/r and common ration = r.

The left hand side is an infinite sum which only converges when the variable is in a certain bound. 

The bound is -1 < r < 1. Otherwise it will grow to infinity very quickly.

Having that in mind, let's use the formula with the contraints -1 < r < 1.

Sum of infinite G.P. = $\dfrac{a}{1 - r}$, where a is the first term and r is the common ratio.

So, 

$\dfrac{\dfrac1r}{1 - r} = \dfrac4r\\ \dfrac{\dfrac1{r} \cdot\color{blue}r\color{black}}{1 - r} = \dfrac{4\color{blue}r\color{black}}r\\ \dfrac1{1 - r}= 4\\ 1 - r = \dfrac14$

I believe you can continue from here.

We factor by grouping on both the numerator and the denominator.

$\quad \dfrac{x^3 - x^2 - 4x + 4}{x^2 + 6x + 8}\\ = \dfrac{(x^3 - x^2) - (4x - 4)}{(x^2 + 4x) + (2x + 8)}\\ = \dfrac{x^2(x - 1) - 4(x - 1)}{x(x + 4) + 2(x + 4)}\\ = \dfrac{(x^2- 4)(x - 1)}{(x + 2)(x + 4)}\\ = \dfrac{\color{red}(x + 2)\color{black}(x - 2)(x - 1)}{\color{red}(x + 2)\color{black}(x + 4)}\\$

Now if I do more than that, that would be spoonfeeding.

Notice the pattern:

3¹ = 3

3² = 9

3³ = 27

3⁴ = 81

3⁵ = 243

The pattern goes forever. The last digits will always be 3, 9, 7, 1, 3, 9, 7, 1, ... 

That means if we have some ways to calculate which one of {3, 9, 7, 1} is the last digit, then we get the answer directly.

Because the pattern is periodic every 4 terms, we know that it is related to the power mod 4.

power mod 4 = 1 gives last digit = 3

power mod 4 = 2 gives last digit = 9

power mod 4 = 3 gives last digit = 7

power mod 4 = 0 gives last digit = 1

We attempt to find $14^{159} \text{ mod } 4$.

$\quad 14^{159} \\\equiv 2^{159}\pmod 4\\\equiv 4^{79} \cdot 2 \pmod 4\\\equiv 0\pmod 4$

That means the required last digit is 1.

By Vieta's formula,

$a + b = -\dfrac4k\\ ab = \dfrac4k$

We start with an identity to find the value of a² + b².

$a^2 + 2ab + b^2 = (a + b)^2\\ a^2 + b^2 + 2\left(\dfrac4k\right) = \left(-\dfrac4k\right)^2\\ a^2 + b^2 = \dfrac{16}{k^2} - \dfrac8k =\dfrac{8(2 - k)}{k^2}$

Then, we equate it to 24.

$\dfrac{8(2-k)}{k^2} = 24\\ 3k^2 = 2 - k\\ 3k^2 + k - 2 = 0$

I believe you can do the rest.

Can you show me what you did to get 4/5?

$z\overline{w} = 6 + 8i\\ \overline{z\overline{w}} = \overline{6+8i}\\ \overline{z}w = 6 - 8i$

And then, we know that for any complex number z and w, $z\overline{z} = |z|^2$ and $|z||w| = |zw|$.

$\quad|zw|^2 \\= |z|^2 |w|^2 \\= z\overline{z}w\overline{w} \\= (z\overline{w})(w\overline z) \\= (6 + 8i)(6 - 8i) \\= 10$

$\quad|z + w|^2 \\= (z + w)\left(\overline{z + w}\right) \\= (z + w)\left(\overline z + \overline w\right)\\ = z\overline z + w\overline w + z\overline w + w\overline z \\= |z|^2 + |w|^2 + z\overline w + \overline zw \\= 5^2 + 2^2 + (6 + 8i) + (6- 8i) \\= 41$

I will leave the remaining ones to you.

$\csc x \tan x = \dfrac1{\sin x} \cdot \dfrac{\sin x}{\cos x} = \dfrac1{\cos x}$

$\dfrac1{\cos x} = \dfrac{2\sqrt 3}3\\ \dfrac1{\cos x} = \dfrac{2}{\sqrt 3}\\ \cos x = \dfrac{\sqrt 3}{2}\\ x = \dfrac{\pi}{6} \text{ or } x = \dfrac{11\pi}{6}$

If the line is tangent to the circle, then the line and the circle only touch at one point. That means if I solve for x( or y), there should only be one solution.

$\begin{cases}x = 2y + 5\\x^2 + y^2 = 5\end{cases}$

Substituting the first equation into the second, 

$(2y + 5)^2 + y^2 = 5\\ 4y^2 + 20y + 25 + y^2 = 5\\ 5y^2 + 20y + 20 = 0\\ y^2 + 4y + 4 = 0\\ (y + 2)^2 = 0\\ \boxed{y = -2}$

As you see, it only has one solution. Therefore the line is indeed tangent to the circle.