$2047$

Akan diberikan dua teknik untuk menyelesaikan soal ini.

Teknik 1

$\begin{array}{ccc}\mathrm{total}& =& 1+2+4+8+\dots +1024\\ \mathrm{total}+1& =& 1+1+2+4+8+\dots +1024\\ \mathrm{total}+1& =& 2+2+4+8+\dots +1024\\ \mathrm{total}+1& =& 4+4+8+\dots +1024\\ \mathrm{total}+1& =& 8+8+\dots +1024\\ \dots \\ \mathrm{total}+1& =& 1024+1024\\ \mathrm{total}+1& =& 2048\\ \mathrm{total}& =& 2047\end{array}$

Teknik 2

$\mathrm{total}=1+2+4+8+\dots +1024$
$=(1+2+4+8+\dots +1024)\times 1$
$=(1+2+4+8+\dots +1024)\times (2-1)$
$=(2+4+8+16+\dots +2048)-(1+2+4+8+\dots +1024)$
$=2048-1$
$=2047$

${\displaystyle \frac{3}{55}}$

${\displaystyle \frac{1}{35}+\frac{1}{63}+\frac{1}{99}}$

$={\displaystyle \frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}}$

$={\displaystyle \frac{1}{2}\cdot (\frac{2}{5\cdot 7}+\frac{2}{7\cdot 9}+\frac{2}{9\cdot 11})}$

$={\displaystyle \frac{1}{2}\cdot (\frac{7-5}{5\cdot 7}+\frac{9-7}{7\cdot 9}+\frac{11-9}{9\cdot 11})}$

$={\displaystyle \frac{1}{2}\cdot (\frac{7}{5\cdot 7}-\frac{5}{5\cdot 7}+\frac{9}{7\cdot 9}-\frac{7}{7\cdot 9}+\frac{11}{9\cdot 11}-\frac{9}{9\cdot 11})}$

$={\displaystyle \frac{1}{2}\cdot (\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})}$

$={\displaystyle \frac{1}{2}\cdot (\frac{1}{5}-\frac{1}{11})}$

$={\displaystyle \frac{1}{2}\cdot \left(\frac{11-5}{5\cdot 11}\right)}$

$={\displaystyle \frac{6}{2\cdot 5\cdot 11}}$

$={\displaystyle \frac{3}{55}}$

${\displaystyle \frac{9}{52}}$

${\displaystyle \frac{3}{28}+\frac{3}{70}+\frac{3}{130}}$

$={\displaystyle \frac{3}{4\mathrm{\cdot }7}+\frac{3}{7\mathrm{\cdot }10}+\frac{3}{10\mathrm{\cdot }13}}$

$={\displaystyle (\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+(\frac{1}{10}-\frac{1}{13})}$

$={\displaystyle \frac{1}{4}-\frac{1}{13}}$

$={\displaystyle \frac{13-4}{4\cdot 13}}$

$={\displaystyle \frac{9}{52}}$

${\displaystyle \frac{1}{2}}$

${\displaystyle \frac{7}{12}-\frac{9}{20}+\frac{11}{30}}$

$={\displaystyle \frac{7}{3\cdot 4}-\frac{9}{4\cdot 5}+\frac{11}{5\cdot 6}}$

$={\displaystyle (\frac{1}{3}+\frac{1}{4})-(\frac{1}{4}+\frac{1}{5})+(\frac{1}{5}+\frac{1}{6})}$

$={\displaystyle \frac{1}{3}+\frac{1}{6}}$

$={\displaystyle \frac{1}{2}}$

${\displaystyle \frac{5}{34}}$

${\displaystyle \frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}}$

$={\displaystyle \frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\frac{1}{11\cdot 14}+\frac{1}{14\cdot 17}}$

$={\displaystyle \frac{1}{3}(\frac{1}{2}-\frac{1}{5})+\frac{1}{3}(\frac{1}{5}-\frac{1}{8})+\frac{1}{3}(\frac{1}{8}-\frac{1}{11})+\frac{1}{3}(\frac{1}{11}-\frac{1}{14})+\frac{1}{3}(\frac{1}{14}-\frac{1}{17})}$

$={\displaystyle \frac{1}{3}(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17})}$

$={\displaystyle \frac{1}{3}(\frac{1}{2}-\frac{1}{17})}$

$={\displaystyle \frac{1}{3}\left(\frac{17-2}{2\cdot 17}\right)}$

$={\displaystyle \frac{15}{3\cdot 2\cdot 17}}$

$={\displaystyle \frac{5}{2\cdot 17}}$

$={\displaystyle \frac{5}{34}}$

$9$

Perhatikan teknik manipulasi di bawah ini.

Untuk $n\in \mathbb{N}$ berlaku:

${\displaystyle \frac{1}{\sqrt{n}+\sqrt{n+1}}}$

$={\displaystyle \frac{1}{(\sqrt{n}+\sqrt{n+1})}\cdot \frac{(\sqrt{n}-\sqrt{n+1})}{(\sqrt{n}-\sqrt{n+1})}}$

$={\displaystyle \frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}}$

$={\displaystyle \frac{\sqrt{n}-\sqrt{n+1}}{\mathrm{-1}}}$

$=\sqrt{n+1}-\sqrt{n}$

Sekarang, identitas baru ini akan digunakan untuk menjawab soal.

${\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots +\frac{1}{\sqrt{99}+\sqrt{100}}}$

$=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\dots +(\sqrt{100}-\sqrt{99})$

$=(\sqrt{2}+\sqrt{3}+\sqrt{4}+\dots +\sqrt{100})-(1+\sqrt{2}+\sqrt{3}+\dots +\sqrt{99})$

$=\sqrt{100}-1$

$=9$

${\displaystyle \frac{1}{3}}$

$x={\displaystyle \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots }$

$4x={\displaystyle 1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots }$

$4x={\displaystyle 1+(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots )}$

$4x=1+x$

$x={\displaystyle \frac{1}{3}}$

$6$

Misalkan:
$a=n^{{\displaystyle \frac{1}{3}}}$
$b={(n-1)}^{{\displaystyle \frac{1}{3}}}$

Maka:

${\displaystyle \frac{1}{\sqrt[3]{n^2}+\sqrt[3]{n(n-1)}+\sqrt[3]{{(n-1)}^2}}}$

$=\frac{1}{a^2+ab+b^2}$

$=\frac{a-b}{(a-b)(a^2+ab+b^2)}$

$=\frac{a-b}{a^3-b^3}$

$=\frac{a-b}{n-(n-1)}$

$=a-b$

$=n^{\frac{1}{3}}-{(n-1)}^{\frac{1}{3}}$

Sekarang

${\displaystyle \sum _{n=1}^{216}{n}^{\frac{1}{3}}-{(n-1)}^{\frac{1}{3}}}$

$={\displaystyle (1^{\frac{1}{3}}-0^{\frac{1}{3}})+(2^{\frac{1}{3}}-1^{\frac{1}{3}})+(3^{\frac{1}{3}}-2^{\frac{1}{3}})+\dots +({216}^{\frac{1}{3}}-{215}^{\frac{1}{3}})}$

${\displaystyle =(1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+\dots +{216}^{\frac{1}{3}})-(0^{\frac{1}{3}}+1^{\frac{1}{3}}+2^{\frac{1}{3}}+\dots +{215}^{\frac{1}{3}})}$

${\displaystyle ={216}^{\frac{1}{3}}}$

$=6$

$2019045$

$S=1^2-2^2+3^2-4^2+\dots +{2007}^2-{2008}^2+{2009}^2$

$S={2009}^2-{2008}^2+{2007}^2-{2006}^2+\dots +5^2-4^2+3^2-2^2+1^2-0^2$

$S=({2009}^2-{2008}^2)+({2007}^2-{2006}^2)+\dots +(5^2-4^2)+(3^2-2^2)+(1^2-0^2)$

$S={\displaystyle \sum _{n=0}^{1004}{(2n+1)}^2-{(2n)}^2}$

$S={\displaystyle \sum _{n=0}^{1004}4n+1}$

$S=4{\displaystyle \sum _{n=0}^{1004}n+\sum _{n=0}^{1004}1}$

$S=4\cdot {\displaystyle \frac{1}{2}\cdot 1004\cdot 1005}+1005$

$S=2019045$

${1.01}^{100}>2$

Latihan yang diberikan pada soal-soal sebelumnya adalah latihan menyederhanakan barisan dan deret. Soal ini adalah kebalikannya. Soal ini dipilih karena bisa men-demonstrasi-kan penggunaan konsep barisan dan deret untuk memberikan analisa yang meyakinkan dan argumen yang kuat untuk menjawab soal ketaksamaan.

${\displaystyle \frac{101}{100}=\frac{101}{100}}\text{ ... (1)}$

Karena ${\displaystyle \frac{1}{100}>\frac{1}{101}}$ maka ${\displaystyle 1+\frac{1}{100}>1+\frac{1}{101}}$ maka ${\displaystyle \frac{101}{100}>\frac{102}{101}}\text{ ... (2)}$

Karena ${\displaystyle \frac{1}{100}>\frac{1}{102}}$ maka ${\displaystyle 1+\frac{1}{100}>1+\frac{1}{102}}$ maka ${\displaystyle \frac{101}{100}>\frac{103}{102}}\text{ ... (3)}$

Karena ${\displaystyle \frac{1}{100}>\frac{1}{103}}$ maka ${\displaystyle 1+\frac{1}{100}>1+\frac{1}{103}}$ maka ${\displaystyle \frac{101}{100}>\frac{104}{103}}\text{ ... (4)}$

......

Karena ${\displaystyle \frac{1}{100}>\frac{1}{199}}$ maka ${\displaystyle 1+\frac{1}{100}>1+\frac{1}{199}}$ maka ${\displaystyle \frac{101}{100}>\frac{200}{199}}\text{ ... (100)}$

Sekarang kalikan (1), (2), (3), (4), ... (100)

${\displaystyle \frac{101}{100}\cdot \frac{101}{100}\cdot \frac{101}{100}\cdot \frac{101}{100}\dots \frac{101}{100}>\frac{101}{100}\cdot \frac{102}{101}\cdot \frac{103}{102}\cdot \frac{104}{103}\dots \frac{200}{199}}$

${\displaystyle {\left(\frac{101}{100}\right)}^{100}>\frac{200}{100}}$

${\displaystyle {1.01}^{100}>2}$