Barisan dan Deret #4 (T.A.M.A.T)

Prinsip Teleskopik

Prinsip teleskopik banyak digunakan untuk menyederhanakan suatu deret. Ada dua bentuk umum yang dikenal, yaitu penjumlahan dan perkalian sebagai berikut.

$latex \sum_{i=1}^{n}(a_{i+1}-a_{1})=(a_{2}-a_{1})+(a_{3}-a_{1})+(a_{4}-a_{1})+...+(a_{n+1}-a_{1})=(a_{n+1}-a_{1})$

$latex \prod_{i=1}^{n}\frac{a_{i+1}}{a_{i}}=\frac{a_{2}}{a_{1}}\times\frac{a_{3}}{a_{2}}\times\frac{a_{4}}{a_{3}}\times...\times\frac{a_{n}}{a_{n-1}}\times\frac{a_{n+1}}{a_{n}}=\frac{a_{n+1}}{a_{1}}$

Contoh 1:

$latex \begin{array}{rcl}\big(1-\frac{1}{3}\big)\big(1-\frac{1}{5}\big)\big(1-\frac{1}{7}\big)...\big(1-\frac{1}{2003}\big)\big(1-\frac{1}{2005}\big)\big(1+\frac{1}{2}\big)\big(1+\frac{1}{4}\big)\big(1+\frac{1}{6}\big)...\big(1+\frac{1}{2004}\big)\big(1+\frac{1}{2006}\big)&=&....\end{array}$

Jawaban:

Misalkan

$latex \begin{array}{rcl}S&=&\big(1-\frac{1}{3}\big)\big(1-\frac{1}{5}\big)\big(1-\frac{1}{7}\big)...\big(1-\frac{1}{2003}\big)\big(1-\frac{1}{2005}\big)\big(1+\frac{1}{2}\big)\big(1+\frac{1}{4}\big)\big(1+\frac{1}{6}\big)...\big(1+\frac{1}{2004}\big)\big(1+\frac{1}{2006}\big)\end{array}$

maka

$latex S=\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}\times...\times\frac{2004}{2005}\times\frac{3}{2}\times\frac{5}{4}\times\frac{7}{6}\times...\times\frac{2007}{2006}$

$latex S=\frac{2}{3}\times\frac{3}{2}\times\frac{4}{5}\times\frac{5}{4}\times\frac{6}{7}\times\frac{7}{6}\times...\times\frac{2004}{2005}\times\frac{2005}{2004}\times\frac{2007}{2006}$

$latex S=\frac{2007}{2006}$

Contoh 2:

Tentukan nilai dari

$latex \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2005.2006}$

Jawaban:

(soal ini merupakan contoh penerapan prinsip teleskopik)

$latex \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2},\ \ \ \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},\ \ \ \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4},\ \ \ ...,\ \ \ \frac{1}{2005.2006}=\frac{1}{2005}-\frac{1}{2006}$

sehingga diperoleh

$latex \begin{array}{rcl}\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2005.2006}&=&\big(\frac{1}{1}-\frac{1}{2}\big)+\big(\frac{1}{2}-\frac{1}{3}\big)+\big(\frac{1}{3}-\frac{1}{4}\big)+\big(\frac{1}{4}-\frac{1}{5}\big)+...+\big(\frac{1}{2005}-\frac{1}{2006}\big)\\\\&=&1-\frac{1}{2006}\\\\&=&\frac{2005}{2006}\end{array}$

Contoh 3:

Buktikan bahwa

$latex \frac{1}{1999}<\frac{1}{2}\ .\ \frac{3}{4}\ .\ \frac{5}{6}\ .\ ...\ .\ \frac{1997}{1998}<\frac{1}{44}$

Jawaban:

Misalkan

$latex P=\frac{1}{2}\ .\ \frac{3}{4}\ .\ \frac{5}{6}\ .\ ...\ .\ \frac{1997}{1998}$  dan $latex Q=\frac{2}{3}\ .\ \frac{4}{5}\ .\ \frac{6}{7}\ .\ ...\ .\ \frac{1998}{1999}$

maka diperoleh

$latex \begin{array}{rcl}PQ&=&\frac{1}{2}\ .\ \frac{3}{4}\ .\ \frac{5}{6}\ .\ ...\ .\ \frac{1997}{1998}.\frac{2}{3}\ .\ \frac{4}{5}\ .\ \frac{6}{7}\ .\ ...\ .\ \frac{1998}{1999}\\\\&=&\frac{1}{2}\ .\ \frac{2}{3}\ .\ \frac{3}{4}\ .\ \frac{4}{5}\ .\ \frac{5}{6}\ .\ \frac{6}{7}\ .\ ...\ .\ \frac{1997}{1998}\ .\ \frac{1998}{1999}\\\\&=&\frac{1}{1999}\end{array}$

Bandingkan secara berurutan sepasang faktor dari$latex P$dan$latex Q$, maka jelaslah$latex P<Q$sehingga diperoleh

$latex P^2<PQ$

$latex P^2<\frac{1}{1999}$

Karena$latex 44^2=1936$maka diperoleh 

$latex P^2<\frac{1}{1999}<\frac{1}{44^2}$
$latex P<\frac{1}{44}$ ........................................... (*)

Bandingkan secara berurutan sepasang faktor dari$latex P$dan$latex \frac{1}{3}\ .\ \frac{3}{5}\ .\ \frac{5}{7}\ .\ ...\ .\ \frac{1997}{1999}=\frac{1}{1999}$, maka jelaslah

$latex P>\frac{1}{1999}$ ........................................... (**)

Berdasarkan (*) dan (**) diperoleh

$latex \frac{1}{1999}<P<\frac{1}{44}$
$latex \frac{1}{1999}<\frac{1}{2}\ .\ \frac{3}{4}\ .\ \frac{5}{6}\ .\ ...\ .\ \frac{1997}{1998}<\frac{1}{44}$

(Terbukti)

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