Soal-Soal Matematika/Barisan dan deret aritmatika

Rumus barisan dan deret aritmatika

${\begin{aligned}u_{n}=a+(n-1)b\\s_{n}={\frac {n(2a+(n-1)b}{2}}\\b=u_{n}-u_{n-1}\\u_{t}={\frac {u_{1}+u_{n}}{2}}\\\end{aligned}}$

suku dan beda baru

$n=n+(n-1)x$
$b_{b}={\frac {b}{x+1}}$

barisan dan deret bertingkat

$b={\frac {a_{n}-a_{k}}{n-k}}$

nb: n dan k adalah suku ke-n dan suku ke-k.

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	$S_{n}=n(n+1)$

• ${\frac {1}{2}}+{\frac {1}{2^{2}}}+{\frac {1}{2^{3}}}+\dots +{\frac {1}{2^{n}}}={\frac {2^{n}-1}{2^{n}}}$
• ${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}=(1-{\frac {1}{2}})+({\frac {1}{2}}-{\frac {1}{3}})+({\frac {1}{3}}-{\frac {1}{4}})=1-{\frac {1}{4}}={\frac {3}{4}}$
• ${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}+{\frac {1}{4\times 5}}={\frac {4}{5}}$
• ${\frac {1}{n\times (n+1)}}={\frac {1}{n}}-{\frac {1}{n+1}}$ (deret teleskop)
• ${\frac {1}{1\times 2}}+{\frac {1}{2\times 3}}+{\frac {1}{3\times 4}}+\dots +{\frac {1}{(n-1)\times n}}={\frac {n-1}{n}}$
• ${\frac {6}{1\times 4\times 7}}+{\frac {6}{4\times 7\times 10}}+{\frac {6}{7\times 10\times 13}}+\dots +{\frac {6}{94\times 97\times 100}}={\frac {1}{1\times 4}}-{\frac {1}{4\times 7}}+{\frac {1}{4\times 7}}-{\frac {1}{7\times 10}}+{\frac {1}{7\times 10}}-{\frac {1}{10\times 13}}+\dots +{\frac {1}{94\times 97}}-{\frac {1}{97\times 100}}={\frac {1}{1\times 4}}-{\frac {1}{97\times 100}}={\frac {1}{4}}-{\frac {1}{9700}}={\frac {2425-1}{9700}}={\frac {2424}{9700}}={\frac {606}{2425}}$
• ${\frac {c-a}{a\times b\times c}}={\frac {1}{a\times b}}-{\frac {1}{b\times c}}$ (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
• ${\frac {1}{4}}+{\frac {11}{44}}+{\frac {111}{444}}={\frac {1}{4}}\times 3={\frac {3}{4}}$
• ${\frac {1}{3}}+{\frac {11}{33}}+{\frac {111}{333}}+\dots +{\frac {111.111}{333.333}}={\frac {1}{3}}\times 6=2$