synthetic division
$ x=16^{\log_23}
$ \log_2x={\log_23}\log_216=4\log_23
$ 6 = 2^1 \times 3^1 なので$ 1 + 2 + 3 + 6 = 2^0 \times 3^0 + 2^1 \times 3^0 + 2^0 \times 3^1 + 2^1 \times 3^1
$ = (2^0 + 2^1)\times(3^0 + 3^1) = (1+2)(1+3) = 12
$ 100 = 2^2 \times 5^2 なので$ 1 + 2 + 4 + 5 + 10 + 20 + 25 + 50 + 100
$ = 2^0 \times 5^0 + 2^1 \times 5^0 + 2^2 \times 5^0 + 2^0 \times 5^1 + 2^1 \times 5^1 + 2^2 \times 5^1 + 2^0 \times 5^2 + 2^1 \times 5^2 + 2^2 \times 5^2
$ = (2^0 + 2^1 + 2^2) \times 5^0 + (2^0 + 2^1 + 2^2) \times 5^1 + (2^0 + 2^1 + 2^2) \times 5^2
$ = (2^0 + 2^1 + 2^2) \times (5^0 + 5^1 + 5^2)
$ = (1+2+4)(1+5+25) = 7 \times 31 = 217
$ 110 = 2^1 \times 5^1 \times 11^1 \longrightarrow \sigma(110) = (1+2)(1+5)(1+11) だめ
$ 120 = 2^3 \times 3^1 \times 5^1 \longrightarrow \sigma(120) = (1+2+4+8)(1+3)(1+5)=15\times4\times6 = 360 いけた
$ 130, 140, 150 いらん(問題に1つだけあるって書いてあるから)
$ \begin{array}{c} x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0\ の解は\ x=a,\, b,\, c\ なので\end{array}
$ \begin{array}{c} a^3 - (a+b+c)a^2 + (ab+bc+ca)a - abc = 0 \\ b^3 - (a+b+c)b^2 + (ab+bc+ca)b - abc = 0 \\ c^3 - (a+b+c)c^2 + (ab+bc+ca)c - abc = 0 \\ \hline (a^3+b^3+c^3) - (a+b+c)(a^2+b^2+c^2-ab-bc-ca)-3abc=0\end{array}
$ k^3 = \frac{k(k+1)(k+2)}{2} + \frac{k(k-1)(k-2)}{2} - 2k
$ \sum k^3 = \sum \frac{k(k+1)(k+2)}{2} + \sum \frac{k(k-1)(k-2)}{2} - 2 \sum k
$ = \frac{N(N+1)(N+2)(N+3)}{8} + \frac{(N+1)N(N-1)(N-2)}{8} - N(N+1)
$ = N(N+1) \left( \frac{(N+2)(N+3)}{8} + \frac{(N-1)(N-2)}{8} - 1 \right)
$ = N(N+1) \frac{N^2+5N+6+N^2-3N+2-8}{8}
$ a_{n+1} = p\, a_{n} + q の解法の中の$ \alpha = p \alpha + q は
$ a_{1} = a_{2} = a_{3} = \cdots = a_{n} = \cdots = \alpha
となる『特別な初項$ \alpha 』(漸化式の不動点)を求めているんだよ!
2項間漸化式のときは、数列を特別な初項の分だけ
$ a_{1}-\alpha,\ a_{2}-\alpha,\ a_{3}-\alpha,\ \cdots,\ a_{n}-\alpha,\ \cdots
とシフトすると等比数列$ a_{n+1} - \alpha = p( a_{n} - \alpha ) になるんだよ!
この考え方は
スクイズの定理(squeeze theorem)
$ a を含む区間で \underbrace{g(x)}_{\rm bunt:突く} < \underbrace{f(x)}_{3塁走者} < \underbrace{h(x)}_{本塁} かつ \lim_{x\to a}g(x)=\lim_{x\to a}h(x)=\alpha
$ \Longrightarrow \underbrace{\lim_{x\to a}g(x)}_{\rm squeeze成功} = \underbrace{\lim_{x\to a}f(x)}_{本塁生還} = \underbrace{\lim_{x\to a}h(x)}_{本塁}=\alpha
例:$ \underbrace{\cos\theta}_{\rm bunt:突く} < \underbrace{\frac{\sin\theta}{\theta}}_{3塁走者} < \underbrace{1}_{本塁} \Longrightarrow \underbrace{\lim_{\theta\to0}\cos\theta}_{\rm squeeze成功} = \underbrace{\lim_{\theta\to0}\frac{\sin\theta}{\theta}}_{本塁生還} = \underbrace{1}_{本塁}
$ x=1 で接するので接線を$ y=r(x) とすると
$ (2x^2+3x+4) - r(x) = q(x)(x-1)^2 なので
$ 2x^2+3x+4 を$ (x-1)^2 でわると
$ \begin{array}{r|r}\begin{array}{r|rrrrrr} & 2 & 3 & 4 \\\hline 2 & & 4 \\ -1 & & & -2 \\\hline & 2 & \hspace{-.8em} || 7 & 2 \end{array} & \therefore 2x^2+3x+4 = \underbrace{2}_{q(x)}(x-1)^2 + \underbrace{7x+2}_{r(x)} \end{array}
多項式の割り算:$ (2x^3-7x^2+8)\div(x^2-4x+3)=2x+1 \ \cdots \ -2x+5 ,
すなわち:$ 2x^3-7x^2+8=(2x+1)(x^2-4x+3)-2x+5 を組立除法で書くと
$ \begin{array}{r|rrrrrr} & 2 & -7 & 0 & 8 \\\hline 4 & & 8 & 4 \\ -3 & & & -6 & -3 \\\hline & 2 & 1 & \hspace{-.8em} || -2 & 5 \end{array}
$ \begin{array}{rl} \begin{array}{r} {2x^3-7x^2+8=}\\ \\ \\ \end{array} & \hspace{-1.5em} \begin{array}{rrrrrrrr} & 2x^3 & \boxed{-8}x^2 & \boxed{+6}x & & = & 2x (x^2-4x+3) \\ & & +x^2 & \boxed{-4}x & \boxed{+3} & & +1(x^2-4x+3) \\ & & & -2x & + 5 & & -2x+5 \end{array} \\ = \ \ & (2x+1)(x^2-4x+3)-2x+5 \end{array}
$ \begin{array}{rrrrrrrr} 2x^3 & -7x^2 & +0x & + 8 & = & 2x^3 -7x^2 +0x +8 \\ -2x^3 & +8x^2 & -6x & & & +2x(-x^2 +4x -3) \\ & -x^2& +4x & -3 & & +1(-x^2+4x-3) \\ & = & -2x & +5 \end{array}
$ 2x^3-7x^2+8=2x(4x-3)-7(4x-3)+8=8x^2-6x-28x+21+8