MaxWong

$m\angle \text{PQR} = m\angle \text{TSR}= 90^{\circ} \\\therefore \text{PQ} \;\;|| \;\;\text{TS}\\ \therefore m\angle P = m\angle T$

$\cos T = \cos P = \dfrac{2}{3}\\ $

$\cos T = \dfrac{\text{ST}}{\text{TR}}\\ \dfrac{2}{3} = \dfrac{6}{x}\\ x = 9$

It is sequence A052918 on OEIS. a₃ is 5(26) + 5 = 135 as the transition formula can be rewritten as $a_{n+1} = 5a_n + a_{n-1}$.

https://oeis.org/A052918 <-- Here's a link to the page about sequence on OEIS.

1. Third option

2

 $f(-2) = 3-(-2)^2 = -1\\ g(-2) = -1\\ \therefore f(x) = g(x) \text{ when } x = -2.$

$\quad\left(\dfrac{f}{g}\right)(x)\\ =\dfrac{4x^2+6x}{2x^2+13x+15}\\ =\dfrac{2x(2x + 3)}{(2x + 3)(x + 5)}\\ =\dfrac{2x}{x+5}$

Oops. Thank you for the correction. 

There are 99 sequences to sum up.

The n-th sequence looks like this:

n(n+1) + (n+1)(n+2) + ... + 99(100).

We can solve this by summation:

$\quad n(n+1 ) + (n+1)(n+2) + \cdots + 99(100)\\ =(1\cdot2 + 2\cdot 3 + ... + 99(100)) - (1\cdot 2 + 2\cdot 3 + ... + n(n-1))\\ =\displaystyle \sum^{99}_{k=1} (k^2 + k) - \sum^{n-1}_{k=1}(k^2 + k)\\ = \dfrac{99(100)(199)}{6} + \dfrac{99(100)}{2} - \dfrac{n(n-1)(2n-1)}{6} - \dfrac{n(n-1)}{2}\\ = -\dfrac{n^3}{3} + \dfrac{n}{3} + 333300$

Now we need to find the sum of this expression from n = 1 to p, where p is some positive integer.

$\displaystyle\sum^{p}_{n=1} \left(\dfrac{-n^3}{3} + \dfrac{n}{3} + 333300\right)\\ = \dfrac{p(-p^3-2p^2+p+3999602)}{12}$

We sum this from p = 1 to p = 100 and we get 1,508,082,510. 

:)

To Guest: The first sequence is (2 + 3 + 4 + ... + 100) + (6 + 8 + 10 + ... + 200) + ... + (98 * 99 + 98*100 + 99*100).

$\tan Y = \dfrac{\text{XZ}}{\text{XY}} = \dfrac{\sqrt{25^2 - 24^2}}{24} = \dfrac{7}{24}$

$x^5y^2 - x^3 y + 5xy^3 = 0\\ 5x^4 y^2 \dfrac{dx}{dy} + 2x^5y-3x^2y\dfrac{dx}{dy} -x^3 +5y^3 \dfrac{dx}{dy} + 15xy^2 = 0\\ (5x^4y^2 -3x^2y+5y^3)\cdot\dfrac{dx}{dy} = x^3-2x^5y-15xy^2 \\ \dfrac{dx}{dy} = \dfrac{x(x^2 - 2x^4y-15y^2)}{y(5x^4y-3x^2+5y^2)}$

Maximum occurs at $x_1 = x_2 = x_3 = \cdots = x_{101} = \dfrac{1}{\sqrt{101}}$.

Maximum value is $100\left(\dfrac{1}{\sqrt{101}}\right)^2 =\dfrac{100}{101}$.

Notes: The equation $x_1^2 + x_2^2 + \cdots + x_{101}^2 = 1$ represents a 101-sphere with radius 1.

:)

$x = \{1,11\} \implies (x-1)(x-11) = 0\implies x^2 - 12 x + 11 = 0\\ x = \{2,10\} \implies (x-2)(x-10) = 0 \implies x^2 - 12 x + 20 = 0\\ x = \{3,9\} \implies (x-3)(x-9) = 0\implies x^2 - 12x + 27 = 0\\ x = \{4,8\} \implies (x-4)(x-8) = 0\implies x^2 - 12x + 32 = 0\\ x = \{5,7\} \implies (x-5)(x-7) = 0\implies x^2 -12x + 35 = 0\\$