MaxWong

Proving the integral to be LHS is more feasible. Thanks for the help, I haven't tried rewriting the right hand side.

Thank you Melody :)

Define 2 linear functions: \(L_1:Ax+By+C=0\\ L_2:Dx+Ey+F=0\).

From the question: The 2 straight lines intersect at the x-axis. \(\Rightarrow \dfrac{C}{A} = \dfrac{F}{D}\)

As C/A = F/D, C = kF and A = kD for some real k. We can rewrite L₂ as: \(L_2:Ax+ny+C=0\).

The equation representing the 2 straight lines is 

\(A^2x^2+(An+AB)xy+Bny^2+2ACx+(BC+Cn)y+C^2=0\).

Define \(n + B \stackrel{\text{def}}{=} P\).

Now the rewritten equation is \(A^2x^2+APxy+Bny^2+2ACx+CPy+C^2=0\).

WLOG, we can assume A > 0.

In this case, A = \(\sqrt6\).

2AC = -24, so C = \(-2\sqrt6\). => b = C² = 24

CP = 3, so P =  \(-\dfrac{\sqrt6}{4}\).

-a = AP = -3/2

So a = 3/2 :D

There we get (a,b) = (3/2, 24). :D

In both diagrams, the tangent line is given. By definition of derivative, the derivatives of the functions at the point of tangency is just the slope of tangent line.

Therefore, by calculating(or approximating? :D) the slope of the tangents, we get \(p'(2) = 2\) and \(q'(2) = \dfrac{-1}{3}\).

And then, we see the graph of p(x) passes through (2,3), and the graph of q(x) passes through (2, -1). So p(2) = 3 and q(2) = -1.

By product rule of differentiation:

\(\quad r'(2)\\ = p(2) q'(2) + q(2) p'(2)\\ = 3\cdot \dfrac{-1}{3} + (-1)\cdot 2\\ = -1 - 2\\ = -3\)

There we go :)

\(\quad\displaystyle \int \sqrt x \left(1-x^2\right)\;\mathbb{d}x\\ \text{Substitute }u^2 = x\text{.}\\ =\displaystyle \int u \left(1-u^4\right) (2u) \;\mathbb{d}u\\ =2 \displaystyle \int \left(u^2 - u^6\right) \;\mathbb{d}u\\ = \dfrac{2u^3}{3} - \dfrac{2u^7}{7} + \mathbf{C}\\ = \dfrac{2x\sqrt x}{3}-\dfrac{2x^3\sqrt x}{7} + \mathbf C\)

Thank you ;)

\(5^{a_{n+1}-a_n} - 1 = \dfrac{1}{n+\frac{2}{3}}, n\geq 1\)

\(k > 1, k\in \mathbb Z^+\) and \(a_k\) is an integer.

Let us do something to the recursive definition of the sequence.

\(\dfrac{5^{a_{n+1}}-5^{a_n}}{5^{a_n}} = \dfrac{1}{n+\frac{2}{3}}\\ 5^{a_{n+1}} = \left(\dfrac{3}{3n+2}+1\right) \cdot 5^{a_n}\\ 5^{a_{n+1}} = \left(\dfrac{3n+5}{3n+2}\right) \cdot 5^{a_n}\\ a_{n+1} = a_n + \log_5\left(\dfrac{3(n+1)+2}{3n+2}\right)\)

You see somehow it is telescoping :P

Now list some values of a_n for small n.

\(a_1 = 1\\ a_2 = 1 + \log_5\left(\dfrac{8}{5}\right) = \log_5(8)\\ a_3 = 1 + \log_5\left(\dfrac{8}{5}\right) + \log_5\left(\dfrac{11}{8}\right)=1+\log_5\left(\dfrac{11}{5}\right)=\log_5(11)\\\)

You observe that: \(a_n = \log_5\left(3n+2\right)\). 

Now we got the explicit formula for a_n. We are halfway done :) (Nice :D)

If a_k is an integer, then \({3k+2}\) must be a power of 5.

Now the question becomes a number theory question: What is the least value of k such that k > 1 and 3k + 2 is a power of 5?

Let \(5^x = 3k+2\).

\(5^x \equiv 2 \pmod 3\)

\(2^x \equiv 2 \pmod 3\)

This congruence holds true when x is an odd integer.

So the least value of k can be obtained by the equation \(5^3 = 3k+2\). (Why not 5¹?)

\(3k + 2 = 125\\ k = \dfrac{123}{3} = 41\)

Voilà!

Let \(f(n) = \log_{10}\left(S(n)\right)\), where S(n) is the sum of elements of the nth row in the Pascal's triangle.

When n = 0, f(n) = log₁₀(1) = 0

So \(\dfrac{f(n)}{\log_{10}2}\) = 0 when n = 0.

When n = 1, f(n) = log₁₀(2)

So \(\dfrac{f(n)}{\log_{10}2}\) = 1 when n = 1.

When n = 2, f(n) = log₁₀(4) = 2log₁₀(2).

So \(\dfrac{f(n)}{\log_{10}2}\) = 2, when n = 2.

We observe the pattern that \(\dfrac{f(n)}{\log_{10}2} = n \;\forall n \in \mathbb N\).

That's OK. Thanks for your attempt to help. 

He has 4 ways to choose a pair of trousers, 7 ways to choose a shirt, and 3 ways to choose a jacket. 

Therefore, answer is \(4\cdot7\cdot3 = 84\).