heureka

Show that $\dbinom{n-1}{r-1} + \dbinom{n-1}{r} = \dbinom {n}{r}$

$\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{n-1}{r-1} } \\ &=& \dfrac{(n-1)!} {(r-1)!\left(n-1-(r-1)\right)!} \\\\ &=& \dfrac{(n-1)!} {(r-1)!(n-1-r+1)!} \\ \\ &=& \dfrac{(n-1)!} {(r-1)!(n-r)!} \\ \\ && \boxed{ (n-1)! = \dfrac{n!}{n}\\\\(r-1)! = \dfrac{r!}{r} } \\\\ \mathbf{ \dbinom{n-1}{r-1} } &=& \mathbf{ \dfrac{n!*r} {n*r!(n-r)!} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{n-1}{r} } \\ &=& \dfrac{(n-1)!} {r!(n-1-r)!} \\\\ &=& \dfrac{(n-1)!} {r!(n-r-1)!} \\ \\ && \boxed{ (n-1)! = \dfrac{n!}{n}\\(n-r-1)! = \dfrac{(n-r)!}{n-r} } \\\\ \mathbf{ \dbinom{n-1}{r} }&=& \mathbf{ \dfrac{n!*(n-r)} {n*r!(n-r)!} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \dbinom{n-1}{r-1} + \dbinom{n-1}{r} \\\\ &=& \dfrac{n!*r} {n*r!(n-r)!} + \dfrac{n!*(n-r)} {n*r!(n-r)!} \\\\ &=& \dfrac{n!}{r!(n-r)!} \left( \dfrac{r+(n-r)}{n} \right) \\\\ &=& \dfrac{n!}{r!(n-r)!} \left( \dfrac{n}{n} \right) \\\\ &=& \dfrac{n!}{r!(n-r)!} \\\\ \dbinom{n-1}{r-1} + \dbinom{n-1}{r}&=& \mathbf{ \dbinom {n}{r} } \\ \hline \end{array}$

There are numbers$A$ and $B$ for which
$\dfrac A{x-1}+\dfrac B{x+1}=\dfrac{x+2}{x^2-1}$
for every number x\neq\pm1.
Find $B$.

$\small{ \begin{array}{|lrcll|} \hline & \dfrac A{x-1}+\dfrac B{x+1} &=& \dfrac{x+2}{x^2-1} \quad | \quad x^2-1=(x-1)(x+1)\\\\ & \dfrac A{x-1}+\dfrac B{x+1} &=& \dfrac{x+2}{(x-1)(x+1)} \quad | \quad *(x-1)(x+1)\\\\ & \mathbf{ A(x+1) + B(x-1) } &=& \mathbf{ x+2 } \\ \hline 1.~ x=-1:& 0 - 2B &=& -1+2 \\ & -2B &=& 1 \\ & \mathbf{B} &=& \mathbf{ -\dfrac{1}{2} } \\ \hline 2.~ x =1:& 2A+ 0 &=& 1+2 \\ & 2A &=& 3 \\ & \mathbf{A} &=& \mathbf{ \dfrac{3}{2} } \\ \hline \end{array} }$

Let a,b, and c be the roots of $24x^3 - 121x^2 + 87x - 8$ = 0
Find$\log_3(a)+\log_3(b)+\log_3(c)$.

Vieta: $-abc = -\dfrac{8}{24}$

$\begin{array}{|rcll|} \hline && \mathbf{\log_3(a)+\log_3(b)+\log_3(c)} \\\\ &=& \log_3(abc) \\\\ &=& \dfrac{\log(abc)}{\log(3)} \quad | \quad -\dfrac{8}{24}=-abc \\\\ &=& \dfrac{\log \left(\dfrac{8}{24} \right)}{\log(3)} \\\\ &=& \dfrac{\log \left(\dfrac{1}{3} \right)}{\log(3)} \\\\ &=& \dfrac{ \log(1)-\log(3) }{\log(3)} \quad | \quad \log(1) = 0\\\\ &=& \dfrac{ -\log(3) }{\log(3)} \\\\ &=& \mathbf{-1} \\ \hline \end{array}$

If $f(c)=\dfrac{3}{2c-3}$,
find $\dfrac{kn^2}{lm}$
when $f^{-1}(c)\times c \times f(c)$
equals the simplified fraction $\dfrac{kc+l}{mc+n}$,
where $k,l,m,\text{ and }n$ are integers.

$\begin{array}{|rcll|} \hline f\left( f^{-1}(c) \right) &=& c \quad | \quad \mathbf{f(c)=\dfrac{3}{2c-3}} \\ \dfrac{3}{2f^{-1}(c)-3} &=& c \\ 3 &=& c\left( 2f^{-1}(c)-3 \right) \\ 3 &=& 2cf^{-1}(c)-3c \\ 2cf^{-1}(c) &=& 3+3c \\ 2cf^{-1}(c) &=& 3(1+c) \\ \mathbf{ f^{-1}(c) } &=& \mathbf{ \dfrac{3(1+c)}{2c} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{ f^{-1}(c)\times c \times f(c) } \\\\ &=& \dfrac{3(1+c)}{2c}\times c \times \dfrac{3}{2c-3} \\ \\ &=& \dfrac{3(1+c)}{2} \times \dfrac{3}{2c-3} \\ \\ &=& \dfrac{9(1+c)}{2(2c-3)} \\\\ &=& \mathbf{ \dfrac{9c+9}{4c-6} } \quad | \quad \text{compare }~\dfrac{kc+l}{mc+n} \\ \hline k&=& 9 \\ l &=& 9 \\ m &=& 4 \\ n &=& -6 \\\\ && \mathbf{ \dfrac{kn^2}{lm} } \\\\ &=& \dfrac{9*(-6)^2}{9*4} \\\\ &=& \dfrac{(-6)^2}{4} \\\\ &=& \dfrac{36}{4} \\\\ &=& \mathbf{9} \\ \hline \end{array}$

Find $a+b+c$, given that $x+y\neq -1$ and
$\begin{align*} ax+by+c&=x+7,\\ a+bx+cy&=2x+6y,\\ ay+b+cx&=4x+y. \end{align*}$

$\small{ \begin{array}{|lrcll|} \hline (1): & ax+by+c &=& x+7 \\ (2): & a+bx+cy &=& 2x+6y \\ (3): & ay+b+cx &=& 4x+y \\ \hline (1)+(2)+(3): & (ax+by+c) \\ & +(a+bx+cy) \\ & +(ay+b+cx) &=& (x+7) \\ &&& +(2x+6y) +(4x+y) \\ & a(1+x+y) \\ & +b(1+x+y)\\ & +c(1+x+y) &=& 7x+7y+7 \\ & (a+b+c)(1+x+y) &=& 7(1+x+y) \quad | \quad :(1+x+y),~ \mathbf{x+y\neq -1} \\ & \mathbf{a+b+c} &=& \mathbf{7} \\ \hline \end{array} }$

Find all complex numbers z such that z^4=-4

Simplify the following
$(2y-1)(4y^{10}+2y^9+4y^8+2y^7 + y^6)$
Express your answer as a polynomial with the degrees of the terms in decreasing order.

$\begin{array}{|rcll|} \hline && \mathbf{(2y-1)(4y^{10}+2y^9+4y^8+2y^7 + y^6)} \\ &=& (2y-1)y^6(4y^4+2y^3+4y^2+2y + 1) \\ &=& y^6(2y-1)(4y^4+2y^3+4y^2+2y + 1) \\ &=& y^6 \left( 2y(4y^4+2y^3+4y^2+2y + 1)-(4y^4+2y^3+4y^2+2y + 1)\right) \\ &=& y^6\left( 8y^5+4y^4+8y^3+4y^2 + 2y -4y^4-2y^3-4y^2-2y - 1 \right) \\ &=& y^6\left( 8y^5+6y^3 - 1 \right) \\ &=& \mathbf{8y^{11}+6y^9-y^6} \\ \hline \end{array}$

There are values A and B such that
$\dfrac{Bx-11}{x^2-7x+10}=\dfrac{A}{x-2}+\dfrac{3}{x-5}$
Find A+B.

$\small{ \begin{array}{|lrcll|} \hline & \dfrac{Bx-11}{x^2-7x+10} &=&\dfrac{A}{x-2}+\dfrac{3}{x-5} \quad | \quad x^2-7x+10 =(x-2)(x-5) \\\\ & \dfrac{Bx-11}{(x-2)(x-5)} &=&\dfrac{A}{x-2}+\dfrac{3}{x-5} \quad | \quad * (x-2)(x-5)\\\\ & Bx-11 &=& A(x-5)+ 3(x-2) \\ & \mathbf{ Bx-A(x-5) } &=& \mathbf{ 3(x-2) + 11 } \\ \hline 1.~ x=0:& 0 +5A &=& -6+11 \\ & 5A &=& 5 \\ & \mathbf{A} &=& \mathbf{1} \\ \hline 2.~ x =5:& 5B - 0 &=& 9+11 \\ & 5B &=& 20 \\ & \mathbf{B} &=& \mathbf{4} \\ \hline \end{array} }$

$\mathbf{A+B = 5}$

Find the sum of the digits of
$777,777,777,777^2 - 222,222,222,223^2$.

Formula: $a^2-b^2 = (a-b)*(a+b)$

$\begin{array}{|rcll|} \hline && \mathbf{ 777,777,777,777^2 - 222,222,222,223^2 } \\ &=& (777,777,777,777 - 222,222,222,223)*(777,777,777,777 + 222,222,222,223) \\ &=& 555,555,555,554*1,000,000,000,000\\ &=& \mathbf{555,555,555,554,000,000,000,000 }\\ \hline \end{array}$

I is the incenter of ABC. We have CE = 12, AE = 9,
and AB = 15.
Find BC.

$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{BC}{AB} } &=& \mathbf{ \dfrac{CE}{AE} } \\\\ \dfrac{BC}{15} &=& \dfrac{12}{9} \\\\ BC &=& \dfrac{15*12}{9} \\\\ \mathbf{BC} &=& \mathbf{20} \\ \hline \end{array}$