数学B

等差数列の一般項

an=a+(n1)da_n = a + (n-1)d

$$ a_n = a + (n-1)d $$

等差数列の和

Sn=12n(a+an)=12n{2a+(n1)d}S_n = \dfrac{1}{2} n(a + a_n) = \dfrac{1}{2} n \{2a + (n-1)d \}

$$ S_n = \dfrac{1}{2} n(a + a_n) = \dfrac{1}{2} n \{2a + (n-1)d \} $$

等比数列の一般項

an=arn1a_n = ar^{n-1}

$$ a_n = ar^{n-1} $$

等比数列の和

Sn={a(1rn)1r=a(rn1)r1(r1のとき)na(r=1のとき)S_n = \begin{cases} \dfrac{a(1-r^n)}{1-r} = \dfrac{a(r^n-1)}{r-1} \quad (r \neq 1 \text{のとき}) \\ na \quad (r = 1 \text{のとき}) \end{cases}

$$ S_n = \begin{cases} \dfrac{a(1-r^n)}{1-r} = \dfrac{a(r^n-1)}{r-1} \quad (r \neq 1 \text{のとき}) \\ na \quad (r = 1 \text{のとき}) \end{cases} $$

和の公式

k=1nc=cn\sum_{k=1}^{n} c = cn

$$ \sum_{k=1}^{n} c = cn $$

k=1nk=12n(n+1)\sum_{k=1}^{n} k = \dfrac{1}{2} n(n+1)

$$ \sum_{k=1}^{n} k = \dfrac{1}{2} n(n+1) $$

k=1nk2=16n(n+1)(2n+1)\sum_{k=1}^{n} k^2 = \dfrac{1}{6} n(n+1)(2n+1)

$$ \sum_{k=1}^{n} k^2 = \dfrac{1}{6} n(n+1)(2n+1) $$

k=1nrk1=1rn1r\sum_{k=1}^{n} r^{k-1} = \dfrac{1-r^n}{1-r}

$$ \sum_{k=1}^{n} r^{k-1} = \dfrac{1-r^n}{1-r} $$

階差数列

bn=an+1anb_n = a_{n+1} - a_{n}

$$ b_n = a_{n+1} - a_{n} $$

部分分数

1n(n+1)=1n1n+1\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}

$$ \dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1} $$

フィボナッチ数列

an=15[(1+52)n(152)n]a_{n} = \dfrac{1}{\sqrt{5}} \left[ \left( \dfrac{ 1 + \sqrt{5} }{2} \right)^{n} - \left( \dfrac{ 1 - \sqrt{5} }{2} \right)^{n} \right]

$$ a_{n} = \dfrac{1}{\sqrt{5}} \left[ \left( \dfrac{ 1 + \sqrt{5} }{2} \right)^{n} - \left( \dfrac{ 1 - \sqrt{5} }{2} \right)^{n} \right] $$

ベクトル

ABundefined\overrightarrow{AB}

$$ \overrightarrow{AB} $$

ベクトルの計算

aundefined+bundefined=bundefined+aundefined\overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} + \overrightarrow{a}

$$ \overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} + \overrightarrow{a} $$

(aundefined+bundefined)+cundefined=aundefined+(bundefined+cundefined)( \overrightarrow{a} + \overrightarrow{b} ) + \overrightarrow{c} = \overrightarrow{a} + ( \overrightarrow{b} + \overrightarrow{c} )

$$ ( \overrightarrow{a} + \overrightarrow{b} ) + \overrightarrow{c} = \overrightarrow{a} + ( \overrightarrow{b} + \overrightarrow{c} ) $$

k(aundefined+bundefined)=kaundefined+kbundefinedk( \overrightarrow{a} + \overrightarrow{b} )= k \overrightarrow{a} + k \overrightarrow{b}

$$ k( \overrightarrow{a} + \overrightarrow{b} )= k \overrightarrow{a} + k \overrightarrow{b} $$

ベクトルの大きさ

aundefined=a12+a22|\overrightarrow{a}| = \sqrt{a_{1}^{2} + a_{2}^{2}}

$$ |\overrightarrow{a}| = \sqrt{a_{1}^{2} + a_{2}^{2}} $$

単位ベクトル

1aundefinedaundefined=(a1a12+a22, a2a12+a22)\dfrac{1}{|\overrightarrow{a}|} \overrightarrow{a} = \left( \dfrac{a_1}{\sqrt{a_{1}^{2} + a_{2}^{2}}}, \ \dfrac{a_2}{\sqrt{a_{1}^{2} + a_{2}^{2}}} \right)

$$ \dfrac{1}{|\overrightarrow{a}|} \overrightarrow{a} = \left( \dfrac{a_1}{\sqrt{a_{1}^{2} + a_{2}^{2}}}, \ \dfrac{a_2}{\sqrt{a_{1}^{2} + a_{2}^{2}}} \right) $$

ベクトルの内積

aundefinedbundefined=aundefinedbundefinedcosθ\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta

$$ \overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta $$

内積と垂直

aundefinedbundefined=0    aundefinedbundefined\overrightarrow{a} \cdot \overrightarrow{b} = 0 \iff \overrightarrow{a} \perp \overrightarrow{b}

$$ \overrightarrow{a} \cdot \overrightarrow{b} = 0 \iff \overrightarrow{a} \perp \overrightarrow{b} $$

内積と成分

aundefinedbundefined=a1b1+a2b2\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2

$$ \overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 $$

三角形の面積

S=12a1b2+a2b1S = \dfrac{1}{2} | a_1 b_2 + a_2 b_1 |

$$ S = \dfrac{1}{2} | a_1 b_2 + a_2 b_1 | $$

直線のベクトル方程式

pundefined=(1t)aundefined+tbundefined\overrightarrow{p} = (1-t) \overrightarrow{a} + t \overrightarrow{b}

$$ \overrightarrow{p} = (1-t) \overrightarrow{a} + t \overrightarrow{b} $$

円のベクトル方程式

pundefinedcundefined=r|\overrightarrow{p} - \overrightarrow{c}| = r

$$ |\overrightarrow{p} - \overrightarrow{c}| = r $$

(pundefinedcundefined)(pundefinedcundefined)=r2(\overrightarrow{p} - \overrightarrow{c}) \cdot (\overrightarrow{p} - \overrightarrow{c}) = r^2

$$ (\overrightarrow{p} - \overrightarrow{c}) \cdot (\overrightarrow{p} - \overrightarrow{c}) = r^2 $$