$\LARGE\color{blue}\textrm{A. Pendahuluan}$
Notasi sigma dari asalnya dari yaitu dari huruf yunani yang memiliki makna jumlah. Dalam matematika lambang notasi sigma $"\sum"$ selanjutnya akan menunjukkan penjumlahan yang teratur sehingga penulisan sebuah deret dari suatu bilangan yang berpola tertentu dapat disederhanakan lebih ringkas.
Sebagai ilustrasinya untuk deretny adalah sebagai berikut
$\begin{array}{ll}\\ \textrm{a}.\quad 1+2+3+4+5+\cdots +100\\ \textrm{b}.\quad 1+3+5+7+9+\cdots +199\\ \textrm{c}.\quad 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+\cdots +100^{2}\end{array}$
Dari bentuk deret di atas jika dimodelkan dengan notasi sigma maka bentuknya akan menjadi lebih sederhana, yaitu:
$\color{blue}\begin{aligned}&\sum_{i=1}^{n}a_{i}=\color{magenta}a_{1}+a_{2}+a_{3}+\cdots +a_{n}\\ &\textrm{Dibaca}:\: \: "\textrm{Jumlah}\: \textrm{dari}\: a_{i}\: \textrm{untuk}\: \: i\\ &\textrm{dari 1 sampai dengan}\: \: n" \: \: \textrm{dan}\: \: a_{i}\: \textrm{adalah suku ke}-i \end{aligned}$
Sehingga contoh ilustrasi deret di atas jika dinotasikan dengan notasi sigma menjadi
$\begin{array}{ll}\\ \textrm{a}.\quad 1+2+3+4+5+\cdots +100=\sum_{i=1}^{100}i\\\\ \textrm{b}.\quad 1+3+5+7+9+\cdots +199=\sum_{i=1}^{100}(2i-1)\\\\ \textrm{c}.\quad 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+\cdots +100^{2}=\sum_{i=1}^{100}i^{2}\end{array}$
$\LARGE\color{blue}\textrm{B. Sifat-Sifat Notasi Sigma}$
Misalkan diketahui $a_{k}$ dan $b_{k}$ adalah suku ke-k dan C adalah sebuah konstanta, maka
$\color{blue}\begin{array}{l}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}C=nC\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}C.a_{k}=C\sum_{k=1}^{n}a_{k}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}\left ( a_{k}+b_{k} \right )=\displaystyle \sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}\left ( a_{k}+b_{k} \right )^{2}=\displaystyle \sum_{k=1}^{n}a_{k}^{2}+2\sum_{k=1}^{n}a_{k}b_{k}+\sum_{k=1}^{n}b_{k}^{2}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n-1}a_{k}+a_{n}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{m}a_{k}+\sum_{k=m+1}^{n}a_{k},\quad 1<m<n \end{array}$
$\LARGE\color{yellow}\fbox{CONTOH SOAL}$
$\begin{array}{ll}\\ 1.&\textrm{Uraikan jumlah berikut dengan lengkap}\\ &\begin{array}{lll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{4}k&\textrm{f}.\quad \displaystyle \sum_{k=1}^{3}2^{k}\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{4}(k-3)&\textrm{g}.\quad \displaystyle \sum_{k=1}^{3}\frac{1}{3^{k}}\\ &\textrm{c}.\quad \displaystyle \sum_{k=1}^{4}5k&\textrm{h}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+1 \right )\\ &\textrm{d}.\quad \displaystyle \sum_{k=1}^{3}(4k+2)&\textrm{i}.\quad \displaystyle \sum_{k=1}^{4}5\left ( \frac{2}{3} \right )^{k}\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{3}(2k+3)&\textrm{j}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+2k-3 \right ) \end{array} \end{array}$
$\begin{aligned}&\textrm{Jawab}\\ &\color{blue}\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah,}\\ &\begin{array}{lll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{4}k=1+2+3+4=10\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{4}(k-3)=(1-3)+(2-3)+(3-3)+(4-3)=-2.\\ &\textrm{c}.\quad \displaystyle \sum_{k=1}^{4}5k=5.1+5.2+5.3+5.4=50\\ &\textrm{d}.\quad \displaystyle \sum_{k=1}^{3}(4k+2)=(4.1+2)+(4.2+2)+(4.3+2)=30\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{3}(2k+3)=(2.1 +3)+(2.2 +3)+(2.3 +3)=21 \\ &\textrm{f}.\quad \displaystyle \sum_{k=1}^{3}2^{k}=2^{1}+2^{2}+2^{3}+2^{4}=2+4+8+16=30\\ &\textrm{g}.\quad \displaystyle \sum_{k=1}^{3}\frac{1}{3^{k}}=\displaystyle \frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}=\displaystyle \frac{9+3+1}{27}=\frac{13}{27}\\ &\textrm{h}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+1 \right )=\cdots +\cdots +\cdots +\cdots +\cdots \\ &\textrm{i}.\quad \displaystyle \sum_{k=1}^{4}5\left ( \frac{2}{3} \right )^{k}=\cdots +\cdots +\cdots +\cdots \\ &\textrm{j}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+2k-3 \right )=\cdots +\cdots +\cdots +\cdots +\cdots \\ \end{array} \end{array} \end{aligned}$
$\begin{array}{ll}\\ 2.&\textrm{Nyatakanlah penjumlahan berikut dengan notasi sigma}\\ &\textrm{a}.\quad 2+4+8+16+32+64\\ &\textrm{b}.\quad 2+6+18+54+162\\ &\textrm{c}.\quad 15+24+35+48\\ &\textrm{d}.\quad \displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}\\ &\textrm{e}.\quad ab+a^{2}b^{2}+a^{3}b^{3}+a^{4}b^{4} \end{array}$
$\color{blue}\begin{aligned}&\color{black}\textrm{Jawab}\\ &(\textrm{a})\quad 2+4+8+16+32+64=\displaystyle \sum_{k=1}^{6}2^{k}\\ &(\textrm{b})\quad 2+6+18+54+162=\displaystyle \sum_{k=1}^{5}2.3^{k-1}\\ &(\textrm{c})\quad 15+24+35+48=\displaystyle \sum_{k=1}^{4}\left ( k^{2}+6k+8 \right )\\ &(\textrm{d})\quad \displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}=\displaystyle \sum_{k=1}^{5}\frac{2^{k}}{(2k+1)}\\ &(\textrm{e})\quad ab+a^{2}b^{2}+a^{3}b^{3}+a^{4}b^{4}=\displaystyle \sum_{k=1}^{4}(ab)^{k}\\ \end{aligned}$
$\begin{array}{ll}\\ 3.&\textrm{Dengan menggunakan kaidah notasi sigma},\\ &\textrm{tunjukkan bahwa}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{6}(2k+3)=2\sum_{k=1}^{6}k+18\\ &\textrm{b}.\quad \displaystyle \sum_{k=3}^{8}(k+3)=\sum_{k=1}^{6}k+30\\ &\textrm{c}.\quad \displaystyle \sum_{k=2}^{5}\left ( 2k^{2}+3k+3 \right )=2\sum_{k=1}^{4}k^{2}+7\sum_{k=1}^{4}k+32\\ &\textrm{d}.\quad \displaystyle \sum_{k=0}^{5}k^{2}=\sum_{k=1}^{6}k^{2}-2\sum_{k=1}^{6}k+6\\ &\textrm{e}.\quad \displaystyle \sum_{k=3}^{6}\left ( k^{2}+2k-3 \right )=\sum_{k=1}^{6}k^{2}+6\sum_{k=1}^{4}+20 \end{array}$
$\color{blue}\begin{aligned}&\color{black}\textrm{Jawab}\\ &(\textrm{a})\quad \displaystyle \sum_{k=1}^{6}(2k+3)=\sum_{k=1}^{6}2k+\sum_{k=1}^{6}3\\ &\: \qquad =\sum_{k=1}^{6}2k+6.3=2\sum_{k=1}^{6}k+18\\ &(\textrm{b})\quad \displaystyle \sum_{k=3}^{8}(k+3)=\sum_{k=3-2}^{8-2}\left ( (k+2)+3 \right )\\ &\: \qquad =\sum_{k=1}^{6}(k+5)=\sum_{k=1}^{6}k+\sum_{k=1}^{6}5=\sum_{k=1}^{6}k+6.5\\ &\: \qquad=\sum_{k=1}^{6}k+30\\ &(\textrm{c})\quad \displaystyle \sum_{k=2}^{5}\left ( 2k^{2}+3k+3 \right )\\ &\: \qquad=\sum_{k=2-1}^{5-1}\left ( 2(k+1)^{2}+3(k+1)+3 \right )=\cdots\\ &(\textrm{d})\quad \displaystyle\sum_{k=0}^{5}k^{2}=\sum_{k=0+1}^{5+1}(k-1)^{2}=\cdots\\ &(\textrm{e})\quad \displaystyle \sum_{k=3}^{6}\left ( k^{2}+2k-3 \right )=\cdots\\ \end{aligned}$
$\LARGE\color{green}\fbox{LATIHAN SOAL}$
$\begin{array}{ll}\\ .&\textrm{Buktikanlah bahwa}\\ &\textrm{a}.\quad \displaystyle \sum_{k=6}^{12}k^{2}=\sum_{k=1}^{7}k^{2}+10\sum_{k=1}^{7}k+175\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{n}(3k-1)^{2}=9\sum_{k=1}^{n}k^{2}-6\sum_{k=1}^{n}k+n\\ &\textrm{c}.\quad \displaystyle \sum_{k=m}^{n}a_{k}=\sum_{k=m+p}^{n+p}a_{k-p}\\ &\textrm{d}.\quad \displaystyle \sum_{i=m}^{n}a_{i}=\sum_{i=1}^{n}a_{i}-\sum_{i=1}^{m-1}a_{i}\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=0}^{n-1}a_{k+1}=\sum_{k=2}^{n+2}a_{k-1}\\ &\textrm{f}.\quad \displaystyle \sum_{k=1}^{n-5}a_{k}=\sum_{k=1}^{n}a_{k}-\sum_{k=(n-5)+1}^{n}a_{k} \end{array}$
DAFTAR PUSTAKA
- Kuntarti, Sulistiyono, Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
- Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: ESIS.
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