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Notasi Sigma dan Induksi Matematika

A. Notasi Sigma
Lambang $"\sum"$ (dibaca Sigma ) adalah huruf besar yunani yang berarti jumlah. Penggunaan notasi sigma di sini adalah untuk mencatat penjumlahan yang teratur, sehingga penulisan suatu deret bilangan yang memiliki pola tertentu dapat dituliskan dengan lebih ringkas.
Perhatikanlah ilustrasi berikut:
$\begin{array}{ll}\\ \textrm{a}.\quad 1+2+3+4+5+\cdots +100\\ \textrm{b}.\quad 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+\cdots +100^{2}\\ \textrm{c}.\quad 0+2+6+12+20+\cdots +380\end{array}$ .
Dari contoh di atas dapat diringkas dengan notasi sigma yang didefinisikan sebagai berikut:
$\begin{aligned}\sum_{i=1}^{n}a_{i}&=a_{1}+a_{2}+a_{3}+\cdots +a_{n} \end{aligned}\\\\ \begin{aligned}\textrm{Dibaca}:\: \: "\textrm{Jumlah}\: \textrm{dari}\: a_{i}\: \textrm{untuk}\: \: i\:\textrm{ dari 1 sampai dengan}\: \: n" \: \: \textrm{dan}\: \: a_{i}\: \textrm{adalah suku ke}-i \end{aligned}$ .
Sehingga untuk no. 1 dapat disingkat sebagai berikut
$1+2+3+4+\cdots +100 =\displaystyle \sum_{i=1}^{100}i$ .
B. Sifat-Sifat Notasi Sigma
$\begin{array}{ll}\\ \textrm{Misalkan}&a_{k}\: \: \textrm{dan}\: \: b_{k}\: \: \textrm{adalah suku ke}-k\: \textrm{dan C adalah suatu konstanta, maka}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}ZE{n}C=nC\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}C.a_{k}=C\sum_{k=1}^{n}a_{k}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}\left ( a_{k}+b_{k} \right )=\displaystyle \sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}\left ( a_{k}+b_{k} \right )^{2}=\displaystyle \sum_{k=1}^{n}a_{k}^{2}+2\sum_{k=1}^{n}a_{k}b_{k}+\sum_{k=1}^{n}b_{k}^{2}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n-1}a_{k}+a_{n}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{m}a_{k}+\sum_{k=m+1}^{n}a_{k},\quad 1<m<n \end{array}$ .
${\fbox{\LARGE{\fbox{CONTOH SOAL}}}}$ .
$\begin{array}{ll}\\ \fbox{1}.&\textrm{Tulislah jumlah berikut dengan lengkap}\\ &\begin{array}{lll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{4}k&\textrm{f}.\quad \displaystyle \sum_{k=1}^{3}2^{k}\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{4}(k-3)&\textrm{g}.\quad \displaystyle \sum_{k=1}^{4}\frac{1}{3^{k}}\\ &\textrm{c}.\quad \displaystyle \sum_{k=1}^{4}5k&\textrm{h}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+1 \right )\\ &\textrm{d}.\quad \displaystyle \sum_{k=1}^{4}(4k+2)&\textrm{i}.\quad \displaystyle \sum_{k=1}^{4}5\left ( \frac{2}{3} \right )^{k}\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{4}(2k+3)&\textrm{j}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+2k-3 \right ) \end{array} \end{array}$ .
Jawab:
$\begin{array}{ll}\\ \fbox{1}.&\textrm{Perhatikanlah,}\\ &\begin{array}{lll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{4}k=1+2+3+4=10\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{4}(k-3)=(1-3)+(2-3)+(3-3)+(4-3)=-2.\\ &\textrm{c}.\quad \displaystyle \sum_{k=1}^{4}5k=5.1+5.2+5.3+5.4=50\\ &\textrm{d}.\quad \displaystyle \sum_{k=1}^{4}(4k+2)=(4.1+2)+(4.2+2)+(4.3+2)+(4.4+2)=48\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{4}(2k+3)=(2.1 +3)+(2\cdots +3)+(2\cdots +3)+(2\cdots +\cdots )=\cdots \\ &\textrm{f}.\quad \displaystyle \sum_{k=1}^{3}2^{k}=2^{1}+2^{2}+2^{3}+2^{4}=2+4+8+16=30\\ &\textrm{g}.\quad \displaystyle \sum_{k=1}^{4}\frac{1}{3^{k}}=\displaystyle \frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}} +\frac{1}{3^{4}}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\displaystyle \frac{27+9+3+1}{81}=\frac{40}{81}\\ &\textrm{h}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+1 \right )=\cdots +\cdots +\cdots +\cdots +\cdots \\ &\textrm{i}.\quad \displaystyle \sum_{k=1}^{4}5\left ( \frac{2}{3} \right )^{k}=\cdots +\cdots +\cdots +\cdots \\ &\textrm{j}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+2k-3 \right )=\cdots +\cdots +\cdots +\cdots +\cdots \\ \end{array} \end{array}$ .
$\begin{array}{ll}\\ \fbox{2}.&\textrm{Nyatakanlah penjumlahan berikut dengan notasi sigma}\\ &\textrm{a}.\quad 2+4+8+16+32+64\\ &\textrm{b}.\quad 2+6+18+54+162\\ &\textrm{c}.\quad 15+24+35+48\\ &\textrm{d}.\quad \displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}\\ &\textrm{e}.\quad ab+a^{2}b^{2}+a^{3}b^{3}+a^{4}b^{4} \end{array}$ .
Jawab:
$\begin{array}{ll}\\ &\textrm{a}.\quad 2+4+8+16+32+64=\displaystyle \sum_{k=1}^{6}2^{k}\\ &\textrm{b}.\quad 2+6+18+54+162=\displaystyle \sum_{k=1}^{5}2.3^{k-1}\\ &\textrm{c}.\quad 15+24+35+48=\displaystyle \sum_{k=1}^{4}\left ( k^{2}+6k+8 \right )\\ &\textrm{d}.\quad \displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}=\displaystyle \sum_{k=1}^{5}\frac{2^{k}}{(2k+1)}\\ &\textrm{e}.\quad ab+a^{2}b^{2}+a^{3}b^{3}+a^{4}b^{4}=\displaystyle \sum_{k=1}^{4}(ab)^{k} \end{array}$ .
$\begin{array}{ll}\\ \fbox{3}.&\textrm{Dengan menggunakan kaidah notasi sigma, tunjukkan bahwa}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{6}(2k+3)=2\sum_{k=1}^{6}k+18\\ &\textrm{b}.\quad \displaystyle \sum_{k=3}^{8}(k+3)=\sum_{k=1}^{6}k+30\\ &\textrm{c}.\quad \displaystyle \sum_{k=2}^{5}\left ( 2k^{2}+3k+3 \right )=2\sum_{k=1}^{4}k^{2}+7\sum_{k=1}^{4}k+32\\ &\textrm{d}.\quad \displaystyle \sum_{k=0}^{5}k^{2}=\sum_{k=1}^{6}k^{2}-2\sum_{k=1}^{6}k+6\\ &\textrm{e}.\quad \displaystyle \sum_{k=3}^{6}\left ( k^{2}+2k-3 \right )=\sum_{k=1}^{6}k^{2}+6\sum_{k=1}^{4}+20 \end{array}$ .
Jawab:
$\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{6}(2k+3)=\sum_{k=1}^{6}2k+\sum_{k=1}^{6}3=\sum_{k=1}^{6}2k+6.3=2\sum_{k=1}^{6}k+18\\ &\textrm{b}.\quad \displaystyle \sum_{k=3}^{8}(k+3)=\sum_{k=3-2}^{8-2}\left ( (k+2)+3 \right )=\sum_{k=1}^{6}(k+5)=\sum_{k=1}^{6}k+\sum_{k=1}^{6}5=\sum_{k=1}^{6}k+6.5=\sum_{k=1}^{6}k+30\\ &\textrm{c}.\quad \displaystyle \sum_{k=2}^{5}\left ( 2k^{2}+3k+3 \right )=\sum_{k=2-1}^{5-1}\left ( 2(k+1)^{2}+3(k+1)+3 \right )=\cdots \\ &\textrm{d}.\quad \displaystyle \sum_{k=0}^{5}k^{2}=\sum_{k=0+1}^{5+1}(k-1)^{2}=\cdots \\ &\textrm{e}.\quad \displaystyle \sum_{k=3}^{6}\left ( k^{2}+2k-3 \right )=\cdots \end{array}$ .
$\begin{array}{ll}\\ \fbox{4}.&\textrm{Buktikanlah bahwa}\\ &\textrm{a}.\quad \displaystyle \sum_{k=6}^{12}k^{2}=\sum_{k=1}^{7}k^{2}+10\sum_{k=1}^{7}k+175\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{n}(3k-1)^{2}=9\sum_{k=1}^{n}k^{2}-6\sum_{k=1}^{n}k+n\\ &\textrm{c}.\quad \displaystyle \sum_{k=m}^{n}a_{k}=\sum_{k=m+p}^{n+p}a_{k-p}\\ &\textrm{d}.\quad \displaystyle \sum_{i=m}^{n}a_{i}=\sum_{i=1}^{n}a_{i}-\sum_{i=1}^{m-1}a_{i}\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=0}^{n-1}a_{k+1}=\sum_{k=2}^{n+2}a_{k-1}\\ &\textrm{f}.\quad \displaystyle \sum_{k=1}^{n-5}a_{k}=\sum_{k=1}^{n}a_{k}-\sum_{k=(n-5)+1}^{n}a_{k} \end{array}$ .

Tutorial Singkat

Rabu, 23 Agustus 2017

Notasi Sigma dan Induksi Matematika

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