Soal-Soal Matematika/Barisan dan deret aritmatika

Rumus barisan dan deret aritmatika

${\begin{aligned}U_{n}&=a+(n-1)b\\S_{n}&={\frac {n(2a+(n-1)b}{2}}\\b&=U_{n}-U_{n-1}\\U_{n}&=S_{n}-S_{n-1}\\U_{t}&={\frac {U_{1}+U_{n}}{2}}\\b&={\frac {U_{n}-U_{k}}{n-k}}\\{\text{nb: n dan k adalah suku ke-n dan suku ke-k.}}\\\end{aligned}}$

keterangan:

a/U₁: suku pertama

n: banyaknya suku ke-n

b: beda suku

Ut: suku tengah

Un: suku ke-n

Sn: jumlah suku ke-n

rataan

$R={\frac {a_{1}+a_{2}+a_{3}+\dots +a_{n}}{n}}$

suku dan beda baru

$n_{b}=n+(n-1)x$

$b_{b}={\frac {b}{x+1}}$

barisan dan deret bertingkat

cara 1

${\begin{aligned}U_{n}&=a+{\frac {(n-1)b}{1!}}+{\frac {(n-1)(n-2)c}{2!}}+\dots \\S_{n}&={\frac {an}{1!}}+{\frac {n(n-1)b}{2!}}+{\frac {n(n-1)(n-2)c}{3!}}+\dots \\\end{aligned}}$

cara 2

• $a_{n}=an^{2}+bn+c$ (tingkat 2)
• $a_{n}=an^{3}+bn^{2}+cn+d$ (tingkat 3)

Rumus istimewa

Bilangan	Suku ke n	Jumlah suku ke n
Bilangan asli	1 + 2 + 3 + 4 + …. + (n-1) + n	$S_{n}={\frac {n(n+1)}{2}}$
Bilangan asli persegi panjang	$1\cdot 2+2\cdot 3+3\cdot 4+4\cdot 5+\dots +n(n+1)$	$S_{n}={\frac {n(n+1)(n+2)}{3}}$
Bilangan asli segitiga	$1+3+6+10+\dots +{\frac {n(n+1)}{2}}$	$S_{n}={\frac {n(n+1)(n+2)}{6}}$
Bilangan asli balok	$1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+4\cdot 5\cdot 6+\dots +n(n+1)(n+2)$	$S_{n}={\frac {n(n+1)(n+2)(n+3)}{4}}$
Kuadrat bilangan asli	$1+2^{2}+3^{2}+4^{2}+\dots +(n-1)^{2}+n^{2}$	$S_{n}={\frac {n(n+1)(2n+1)}{6}}$
Kubik bilangan asli	$1+2^{3}+3^{3}+4^{3}+\dots +(n-1)^{3}+n^{3}$	$S_{n}=({\frac {n(n+1)}{2}})^{2}$
Bilangan ganjil	1 + 3 + 5 + 7 + …. + 2n + (2n-1)	$S_{n}=n^{2}$
Bilangan genap	2 + 4 + 6 + 8 + … + 2n-2 + 2n	$S_{n}=n(n+1)$

• ${\frac {1}{2}}+{\frac {1}{2^{2}}}+{\frac {1}{2^{3}}}+\dots +{\frac {1}{2^{n}}}=1-{\frac {1}{2^{n}}}$
• ${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}=(1-{\frac {1}{2}})+({\frac {1}{2}}-{\frac {1}{3}})+({\frac {1}{3}}-{\frac {1}{4}})=1-{\frac {1}{4}}={\frac {3}{4}}$
• ${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}+{\frac {1}{4\times 5}}={\frac {4}{5}}$
• ${\frac {1}{n\times (n+1)}}={\frac {1}{n}}-{\frac {1}{n+1}}$ (deret teleskop)
• ${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}+\dots +{\frac {1}{(n-1)\times n}}={\frac {n-1}{n}}$
• ${\frac {1}{6\times 10}}+{\frac {1}{10\times 14}}+{\frac {1}{14\times 18}}+{\frac {1}{18\times 22}}={\frac {1}{6\times (6+4)}}+{\frac {1}{10\times (10+4)}}+{\frac {1}{14\times (14+4)}}+{\frac {1}{18\times (18+4)}}={\frac {1}{4}}({\frac {1}{6}}-{\frac {1}{6+4}})+{\frac {1}{4}}({\frac {1}{10}}-{\frac {1}{10+4}})+{\frac {1}{4}}({\frac {1}{14}}-{\frac {1}{14+4}})+{\frac {1}{4}}({\frac {1}{18}}-{\frac {1}{18+4}})={\frac {1}{4}}({\frac {1}{6}}-{\frac {1}{10}}+{\frac {1}{10}}-{\frac {1}{14}}+{\frac {1}{14}}-{\frac {1}{18}}+{\frac {1}{18}}-{\frac {1}{22}})={\frac {1}{4}}({\frac {1}{6}}-{\frac {1}{22}})={\frac {1}{33}}$
• ${\frac {1}{n\times (n+k)}}={\frac {1}{k}}({\frac {1}{n}}-{\frac {1}{n+k}})$ (deret teleskop)

• ${\frac {6}{1\times 4\times 7}}+{\frac {6}{4\times 7\times 10}}+{\frac {6}{7\times 10\times 13}}+\dots +{\frac {6}{94\times 97\times 100}}={\frac {1}{1\times 4}}-{\frac {1}{4\times 7}}+{\frac {1}{4\times 7}}-{\frac {1}{7\times 10}}+{\frac {1}{7\times 10}}-{\frac {1}{10\times 13}}+\dots +{\frac {1}{94\times 97}}-{\frac {1}{97\times 100}}={\frac {1}{1\times 4}}-{\frac {1}{97\times 100}}={\frac {1}{4}}-{\frac {1}{9700}}={\frac {2425-1}{9700}}={\frac {2424}{9700}}={\frac {606}{2425}}$
• ${\frac {c-a}{a\times b\times c}}={\frac {1}{a\times b}}-{\frac {1}{b\times c}}$ (a, b dan c adalah bilangan yang berurutan dengan selisih tetap pada tingkat pertama)
• ${\frac {1}{1\times 2}}-{\frac {1}{2\times 3}}-{\frac {1}{3\times 4}}=(1-{\frac {1}{2}})-({\frac {1}{2}}-{\frac {1}{3}})-({\frac {1}{3}}-{\frac {1}{4}})={\frac {1}{4}}$
• ${\frac {1}{1\times 2}}-{\frac {1}{2\times 3}}-{\frac {1}{3\times 4}}-{\frac {1}{4\times 5}}={\frac {1}{5}}$
• ${\frac {1}{1\times 2}}-{\frac {1}{2\times 3}}-{\frac {1}{3\times 4}}-\dots -{\frac {1}{(n-1)\times n}}={\frac {1}{n}}$
• ${\frac {1}{1+{\sqrt {2}}}}+{\frac {1}{{\sqrt {2}}+{\sqrt {3}}}}+{\frac {1}{{\sqrt {3}}+{\sqrt {4}}}}+{\frac {1}{{\sqrt {4}}+{\sqrt {5}}}}={\sqrt {5}}-1$
• ${\frac {1}{1+{\sqrt {2}}}}+{\frac {1}{{\sqrt {2}}+{\sqrt {3}}}}+{\frac {1}{{\sqrt {3}}+{\sqrt {4}}}}+{\frac {1}{{\sqrt {4}}+{\sqrt {5}}}}+{\frac {1}{{\sqrt {5}}+{\sqrt {6}}}}={\sqrt {6}}-1$
• ${\frac {1}{1+{\sqrt {2}}}}+{\frac {1}{{\sqrt {2}}+{\sqrt {3}}}}+\dots +{\frac {1}{{\sqrt {n-1}}+{\sqrt {n}}}}={\sqrt {n}}-1$
• 1 + 11 + 111 = 123 x 1 = 123
• 2 + 22 + 222 = 123 x 2 = 246
• 3 + 33 + 333 = 123 x 3 = 369
• 1 + 11 + 111 + 1111 = 1234 x 1 = 1234
• 2 + 22 + 222 + 2222 = 1234 x 2 = 2468
• 3 + 33 + 333 + 3333 = 1234 x 3 = 3702
• ${\frac {1}{4}}+{\frac {11}{44}}+{\frac {111}{444}}={\frac {1}{4}}\times 3={\frac {3}{4}}$
• ${\frac {1}{3}}+{\frac {11}{33}}+{\frac {111}{333}}+\dots +{\frac {111.111}{333.333}}={\frac {1}{3}}\times 6=2$

contoh soal

1. Kursi pada gedung bioskop memiliki 10 baris, yang kapasitasnya membentuk barisan aritmatika. baris terdepan 15 kursi dan terbelakang 33 kursi maka berapa kapasitas kursi pada bioskop tersebut?

Jawaban

$${\begin{aligned}n&=10\\a&=15\\U_{n}&=U_{10}=33\\b&={\frac {33-15}{10-1}}\\&={\frac {18}{9}}\\&=2\\S_{10}&={\frac {n}{2}}(a+U_{n})\\&={\frac {10}{2}}(15+33)\\&=5(48)\\&=240\\{\text{jadi kapasitas kursi pada bioskop adalah 240 kursi }}\\\end{aligned}}$$