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the numbers in the sum seem to follow a linear relationship. A way to write this in sigma notation is $\sum_{n=0}^5 4+3n$, but the function inside the sigma could be anything that satisfies this sequence.

1/2

because for every WX > YZ, there is another choice of points where WX < YZ (by symmetry)

The sequence $a_n$ has the property that $a_n = a_{n-1} + 2a_{n-2}$ for $n \ge 2$. It is also true that $a_0 = 4$ and $a_3 = 6$. What is the value of $a_5$? Express your answer as a common fraction.

For n = 3 and n = 2:

$a_3 = a_2 + 2 a_1\\ a_2 = a_1 + 2a_0$

substituting,

$6=a_2+2a_1\\ a_2 = a_1+8\\ $

this is a simple system of equations.

we can find $a_n$ for any n now by plugging $a_{n-1}$ and $a_{n-2}$ into the formula.

For example, $a_4 = a_3 + 2a_2$ (which we know already)

g(x) will reflect f(x) across the x axis, intersecting with f(x) at the 2 roots.

h(x) will reflect f(x) across the y axis, intersecting with f(x) at the y interception of f(x).

to see why g(x) and h(x) are reflections of f(x) across the x and y axis respectively, try seeing what happens when you change the input for the 2 functions.

thanks guest, my answer is very far off lol

$\frac{k(k-1)}{2}$ is really just the sum of integers from 1 to k-1, and the cosine of some integer multiple $\pi$ oscillates between -1 and 1 (if the integer is even, it's 1, and if it's odd, it's -1). Here, the sum alternates 2 negatives and 2 positives (starting with k=19)

This means that the first 4 terms of the sum are

$-(1+2+3+\cdots+18)-(1+2+3+\cdots+19)+(1+2+3+\cdots+20)+(1+2+3+\cdots+21) \\=20+19+20+21 \\=4(20)$

(notice a bunch of cancellation happens between the sums, as the 1+2+3+...18 is repeated throughout all the 4 terms)

So for every group of 4 terms, the sum of the group is the biggest number of the sum of the 3rd term of that group times 4.

Therefore, the sum is:

$4(20+24+28+\cdots+96)$

Use the sum formula to obtain your answer.

differentiating with respect to x (using chain rule) gets:

$2(x+12)+2(x+7)+2(x+3)+2(x-14)+2(x-8)\\ =2(5x)\\ =10x$

There is only 1 critical point $x = 0$, and since this function is concave up (the double derivative is positive), it is a minimum point. Plug in zero to your equation to get your answer.

the limit diverges

to see why, as n approaches infinity, the terms approach 5/4, which is nonzero and therefore diverges.

Modulus just means the length of the complex number from zero.

The values of  | m + n | = 3√10 would sit on a circle with radius 3√10

The best way to reach that radius whilst minimizing the length is to draw a straight line. That is if you represent m and n as vectors, they would have the same direction. 

This could be proven to be the minimum length by using theorems like the law of cosines or triangle inequality.

To illustrate my point, move around n and try to find the minimum length of the black line:

Thanks melody! I suspected that the solution would be simpler than I thought. 

hmm

$a^2+2ab+b^2=6ab\\ (a+b)^2=6ab\\ a+b=\sqrt{6ab}\\ a^2-2ab+b^2=2ab\\ (a-b)^2=2ab\\ a-b=\sqrt{2ab}$

So

 $\frac{a^2-b^2}{ab}\\ =\frac{\sqrt{6ab}\cdot\sqrt{2ab}}{ab}\\ =\sqrt{12}\cdot\frac{ab}{ab}\\ =2\sqrt{3}$

(notice that a and b are both positive so the answer must be positive)

Also, this question was asked quite a while ago:

Thanks melody, I didn't realize the whole thing reduced to 3 haha

Also, if you were to go to WolframAlpha and try to minimize the function, it gives the correct minimum value but at a strangely specific point:

You could apply AM-GM to get

$\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\ge3 \sqrt[3]{\frac{(y-x)^2}{(y-z)(z-x)} \cdot\frac{(z-y)^2}{(z-x)(x-y)} \cdot\frac{(x-z)^2}{(x-y)(y-z)}}\\=\boxed{3}$

but idk if that's correct because it says real numbers not positive reals

Nice answer proyaop, but squaring either one of those doesn't end up with 2i+2

$2+2i\\ =2\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ =2\sqrt{2}(\cos(45^\circ)+i\sin(45^\circ))$

when taking the square root, the argument divides by 2 and the magnitude square roots, so it is:

$=2^{\frac{3}{4}}(\cos(22.5^\circ)+i\sin(22.5^\circ))$

To solve for $\cos(22.5^\circ)$ and $\sin(22.5^\circ)$, we can make use of the half-angle identity:

$\cos(22.5^\circ)\\= \sqrt{\frac{1+\cos(45^\circ)}{2}}\\= \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}\\= \sqrt{\frac{2+\sqrt{2}}{4}}\\= \frac{\sqrt{2+\sqrt{2}}}{2}$

(it's positive because it's in the first quadrant)

Same thing with sin:

$\sin(22.5^\circ)\\= \sqrt{\frac{1-\cos(45^\circ)}{2}}\\= \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\\= \sqrt{\frac{2-\sqrt{2}}{4}}\\= \frac{\sqrt{2-\sqrt{2}}}{2}$

Thus, the final root is:

$2^{\frac{3}{4}}(\frac{\sqrt{2+\sqrt{2}}}{2}+\frac{\sqrt{2-\sqrt{2}}}{2}i)\\ =\frac{\sqrt{2+\sqrt{2}}}{\sqrt[4]{2}}+\frac{\sqrt{2-\sqrt{2}}}{\sqrt[4]{2}}i$

The simplifying is up to you.

Also, there is one more root, if you use $\cos(22.5^\circ+180^\circ)$ and $\sin(22.5^\circ+180^\circ)i$, because there are 2 roots of unity. Notice that the cosine and sine flip signs in that case, because they get to the 3rd quadrant. So the other root is $=-\frac{\sqrt{2+\sqrt{2}}}{\sqrt[4]{2}}-\frac{\sqrt{2-\sqrt{2}}}{\sqrt[4]{2}}i$

I seriously doubt that the problem was meant to be this complicated

Nice, guest, just dropping a non-computation-heavy solution

If a is 1 mod 7, then raising it to any power will still be 1 mod 7, meaning that a^81 is 1 mod 7.

For b, notice that 2^3 is one more than 7, so b^x will be 1 mod 7 for x being multiples of 3. Since 91 is 1 more than a multiple of 3, it will multiply another 2, meaning that b^91 is 2 mod 7.

Lastly, since c is -1 mod 7, c^2 is (-1)*(-1) = 1 mod 7.

Lastly, just multiply all the results together to get 1*2*1=2