HOTS Zone : Deret Teleskopik

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Berikut ini adalah kumpulan soal mengenai Deret Teleskopik. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

STANDARSNBTHOTS

No. 1

Hasil dari ${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{12}}+\cdots +{\displaystyle \frac{1}{110}}$ adalah ....

Evaluasi GMOM Oktober 2023

ALTERNATIF PENYELESAIAN

0

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

$\begin{array}{rl}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{12}}+\cdots +{\displaystyle \frac{1}{110}}& ={\displaystyle \frac{1}{1\times 2}}+{\displaystyle \frac{1}{2\times 3}}+{\displaystyle \frac{1}{3\times 4}}+\cdots +{\displaystyle \frac{1}{10\times 11}}\\ & ={\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}+\cdots +{\displaystyle \frac{1}{10}}-{\displaystyle \frac{1}{11}}\\ & ={\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{11}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{10}{11}}}}}}\end{array}$

Jadi, ${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{12}}+\cdots +{\displaystyle \frac{1}{110}}={\displaystyle \frac{10}{11}}$.

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No. 2

Misal $$T=\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\cdots +\frac{1}{\sqrt{2022}+\sqrt{2023}}$$ maka nilai ⌊T⌋ adalah ....

KOSCO 2023

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{1}{\sqrt{n}+\sqrt{n+1}}}& ={\displaystyle \frac{1}{\sqrt{n+1}+\sqrt{n}}}\text{color}red\cdot {\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}}\\ & ={\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}}\\ & ={\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{1}}\\ & =\sqrt{n+1}-\sqrt{n}\end{array}$

$\begin{array}{rl}T& ={\displaystyle \frac{1}{2+\sqrt{5}}}+{\displaystyle \frac{1}{\sqrt{5}+\sqrt{6}}}+\cdots +{\displaystyle \frac{1}{\sqrt{2022}+\sqrt{2023}}}\\ & =\sqrt{5}-2+\sqrt{6}-\sqrt{5}+\cdots +\sqrt{2023}-\sqrt{2022}\\ & =\sqrt{2023}-2\\ & =44,...-2\\ & =42,...\\ \lfloor T\rfloor & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black42}}}}\end{array}$

Jadi, ⌊T⌋ = 42.

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No. 3

Sederhanakan $\dfrac{2\times3}{1\times4}+\dfrac{5\times6}{4\times7}+\dfrac{8\times9}{7\times10}+\cdots+\dfrac{98\times99}{97\times100}$

Om T

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{(n+1)(n+2)}{n(n+3)}}& ={\displaystyle \frac{n^2+3n+2}{n^2+3n}}\\ & =1+{\displaystyle \frac{2}{n(n+3)}}\\ & =1+{\displaystyle \frac{2}{3}}({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+3}})\end{array}$

$\begin{array}{rl}{\displaystyle \frac{2\times 3}{1\times 4}}+{\displaystyle \frac{5\times 6}{4\times 7}}+{\displaystyle \frac{8\times 9}{7\times 10}}+\cdots +{\displaystyle \frac{98\times 99}{97\times 100}}& =({\displaystyle \frac{97-1}{4-1}}+1)+{\displaystyle \frac{2}{3}}({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{7}}+\cdots +{\displaystyle \frac{1}{99}}-{\displaystyle \frac{1}{100}})\\ & =33+{\displaystyle \frac{2}{3}}({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{100}})\\ & =33+{\displaystyle \frac{2}{3}}\left({\displaystyle \frac{99}{100}}\right)\\ & =33+{\displaystyle \frac{33}{50}}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{\displaystyle \frac{1683}{50}}}}}}\end{array}$

Jadi, $\dfrac{2\times3}{1\times4}+\dfrac{5\times6}{4\times7}+\dfrac{8\times9}{7\times10}+\cdots+\dfrac{98\times99}{97\times100}=\dfrac{1683}{50}$.

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No. 4

Diberikan $x=101!\left(\dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\cdots+\dfrac{99}{100!}\right)$. Tentukan nilai dari 101! − x

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{n}{(n+1)!}}& ={\displaystyle \frac{n+1-1}{(n+1)!}}\\ & ={\displaystyle \frac{n+1}{(n+1)!}}-{\displaystyle \frac{1}{(n+1)!}}\\ & ={\displaystyle \frac{n+1}{(n+1)\cdot n!}}-{\displaystyle \frac{1}{(n+1)!}}\\ & ={\displaystyle \frac{1}{n!}}-{\displaystyle \frac{1}{(n+1)!}}\end{array}$

$\begin{array}{rl}{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{2}{3!}}+{\displaystyle \frac{3}{4!}}+\cdots +{\displaystyle \frac{99}{100!}}& ={\displaystyle \frac{1}{1!}}-{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{3!}}-{\displaystyle \frac{1}{4!}}+\cdots +{\displaystyle \frac{1}{99!}}-{\displaystyle \frac{1}{100!}}\\ & ={\displaystyle \frac{1}{1!}}-{\displaystyle \frac{1}{100!}}\\ & =1-{\displaystyle \frac{1}{100!}}\end{array}$

$\begin{array}{rl}x& =101!(1-{\displaystyle \frac{1}{100!}})\\ & =101!-101\end{array}$

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$\begin{array}{rl}101!-x& =101!-(101!-101)\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black101}}}}\end{array}$

Jadi, 101! − x = 101.

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No. 5

Nilai dari $$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots +\frac{2024}{2025!}$$ dapat dinyatakan sebagai $a-\frac{1}{b!}$ di mana a dan b merupakan bilangan bulat positif. Berapa nilai dari a + b?

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{n}{(n+1)!}}& ={\displaystyle \frac{n+1-1}{(n+1)!}}\\ & ={\displaystyle \frac{n+1}{(n+1)!}}-{\displaystyle \frac{1}{(n+1)!}}\\ & ={\displaystyle \frac{n+1}{(n+1)\cdot n!}}-{\displaystyle \frac{1}{(n+1)!}}\\ & ={\displaystyle \frac{1}{n!}}-{\displaystyle \frac{1}{(n+1)!}}\end{array}$

$\begin{array}{rl}{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{2}{3!}}+{\displaystyle \frac{3}{4!}}+\cdots +{\displaystyle \frac{2024}{2025!}}& ={\displaystyle \frac{1}{1!}}-{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{3!}}-{\displaystyle \frac{1}{4!}}+\cdots +{\displaystyle \frac{1}{2024!}}-{\displaystyle \frac{1}{2025!}}\\ & ={\displaystyle \frac{1}{1!}}-{\displaystyle \frac{1}{2025!}}\\ & =1-{\displaystyle \frac{1}{2025!}}\end{array}$

1 + 2025 = 2026

Jadi, a + b = 2026.

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No. 6

Diketahui $S=1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^3}}+\cdots +{\displaystyle \frac{1}{2^{10}}}$. Tentukan nilai dari ${\displaystyle \frac{S}{2-S}}$.

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}S& =1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^3}}+\cdots +{\displaystyle \frac{1}{2^{10}}}\\ & =1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}+\cdots +{\displaystyle \frac{1}{1024}}\\ & ={\displaystyle \frac{1023}{1024}}\end{array}$

$\begin{array}{rl}{\displaystyle \frac{S}{2-S}}& ={\displaystyle \frac{{\displaystyle \frac{1023}{1024}}}{2-{\displaystyle \frac{1023}{1024}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1023}{1024}}}{{\displaystyle \frac{2048-1023}{1024}}}}\\ & ={\displaystyle \frac{1023}{1025}}\\ & ={\displaystyle \frac{99}{100}}\end{array}$

Jadi, ${\displaystyle \frac{S}{2-S}}={\displaystyle \frac{99}{100}}$.

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No. 7

Hitunglah nilai dari $$\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\cdots +\frac{1}{({2014}^2-1)\sqrt{{2014}^2}+{2014}^2\sqrt{{2014}^2-1}}$$

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}}& ={\displaystyle \frac{1}{{\left(\sqrt{n}\right)}^2\sqrt{n+1}+{\left(\sqrt{n+1}\right)}^2\sqrt{n}}}\\ & ={\displaystyle \frac{1}{\sqrt{n}\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}}\text{color}red\cdot {\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}}\\ & ={\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}(n+1-n)}}\\ & ={\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}}\\ & ={\displaystyle \frac{1}{\sqrt{n}}}-{\displaystyle \frac{1}{\sqrt{n+1}}}\end{array}$

$\begin{array}{rl}{\displaystyle \frac{1}{1\sqrt{2}+2\sqrt{1}}}+{\displaystyle \frac{1}{2\sqrt{3}+3\sqrt{2}}}+\cdots +{\displaystyle \frac{1}{({2014}^2-1)\sqrt{{2014}^2}+{2014}^2\sqrt{{2014}^2-1}}}& =({\displaystyle \frac{1}{\sqrt{1}}}-{\displaystyle \frac{1}{\sqrt{2}}})+({\displaystyle \frac{1}{\sqrt{2}}}-{\displaystyle \frac{1}{\sqrt{3}}})+\cdots +({\displaystyle \frac{1}{\sqrt{{2014}^2-1}}}-{\displaystyle \frac{1}{\sqrt{{2014}^2}}})\\ & =1-{\displaystyle \frac{1}{\sqrt{{2014}^2}}}\\ & =1-{\displaystyle \frac{1}{2014}}\\ & ={\displaystyle \frac{2013}{2014}}\end{array}$

Jadi, $\frac1{1\sqrt2+2\sqrt1}+\frac1{2\sqrt3+3\sqrt2}+\cdots+\frac1{\left(2014^2-1\right)\sqrt{2014^2}+2014^2\sqrt{2014^2-1}}=\dfrac{2013}{2014}$.

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