Soal-Soal Matematika/Barisan dan deret aritmatika

Rumus barisan dan deret aritmatika

$\begin{array}{c}{u}_n=a+(n-1)b\\ s_n=\frac{n(2a+(n-1)b}{2}\\ b=u_n-u_{n-1}\\ u_t=\frac{u_1+u_n}{2}\\ \end{array}$

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

$\begin{array}{c}{u}_n=a+\frac{(n-1)b}{1!}+\frac{(n-1)(n-2)c}{2!}+\dots \\ s_n=\frac{an}{1!}+\frac{n(n-1)b}{2!}+\frac{n(n-1)(n-2)c}{3!}+\dots \\ \end{array}$

cara 2

• $a_n=a{n}^2+bn+c$ (tingkat 2)
• $a_n=a{n}^3+b{n}^2+cn+d$ (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)
• $\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\dots +\frac{1}{(n-1)\times n}=\frac{n-1}{n}$
• $\frac{1}{6\times 10}+\frac{1}{10\times 14}+\frac{1}{14\times 18}+\frac{1}{18\times 22}=\frac{1}{6\times (6+4)}+\frac{1}{10\times (10+4)}+\frac{1}{14\times (14+4)}+\frac{1}{18\times (18+4)}=\frac{1}{4}(\frac{1}{6}-\frac{1}{6+4})+\frac{1}{4}(\frac{1}{10}-\frac{1}{10+4})+\frac{1}{4}(\frac{1}{14}-\frac{1}{14+4})+\frac{1}{4}(\frac{1}{18}-\frac{1}{18+4})=\frac{1}{4}(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22})=\frac{1}{4}(\frac{1}{6}-\frac{1}{22})=\frac{1}{33}$
• $\frac{1}{n\times (n+k)}=\frac{1}{k}(\frac{1}{n}-\frac{1}{n+k})$ (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
• $\frac{1}{4}+\frac{11}{44}+\frac{111}{444}=\frac{1}{4}\times 3=\frac{3}{4}$
• $\frac{1}{3}+\frac{11}{33}+\frac{111}{333}+\dots +\frac{111.111}{333.333}=\frac{1}{3}\times 6=2$