HOTS Zone : Deret Teleskopik
Table of Contents
Tipe:
No.
Hasil dari \(\dfrac12+\dfrac16+\dfrac1{12}+\cdots+\dfrac1{110}\) adalah ....ALTERNATIF PENYELESAIAN
0
\(\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}\)
\(\begin{aligned} \dfrac12+\dfrac16+\dfrac1{12}+\cdots+\dfrac1{110}&=\dfrac1{1\times2}+\dfrac1{2\times3}+\dfrac1{3\times4}+\cdots+\dfrac1{10\times11}\\[3.8pt] &=\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\cdots+\dfrac1{10}-\dfrac1{11}\\[3.8pt] &=\dfrac11-\dfrac1{11}\\ &=\boxed{\boxed{\dfrac{10}{11}}} \end{aligned}\)
\(\begin{aligned} \dfrac12+\dfrac16+\dfrac1{12}+\cdots+\dfrac1{110}&=\dfrac1{1\times2}+\dfrac1{2\times3}+\dfrac1{3\times4}+\cdots+\dfrac1{10\times11}\\[3.8pt] &=\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\cdots+\dfrac1{10}-\dfrac1{11}\\[3.8pt] &=\dfrac11-\dfrac1{11}\\ &=\boxed{\boxed{\dfrac{10}{11}}} \end{aligned}\)
Jadi, \(\dfrac12+\dfrac16+\dfrac1{12}+\cdots+\dfrac1{110}=\dfrac{10}{11}\).
No.
Misal $$T=\frac1{2+\sqrt5}+\frac1{\sqrt5+\sqrt6}+\cdots+\frac1{\sqrt{2022}+\sqrt{2023}}$$ maka nilai ⌊T⌋ adalah ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac1{\sqrt{n}+\sqrt{n+1}}&=\dfrac1{\sqrt{n+1}+\sqrt{n}}{\color{red}\cdot\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}}\\[4pt]
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\\[4pt]
&=\dfrac{\sqrt{n+1}-\sqrt{n}}1\\[4pt]
&=\sqrt{n+1}-\sqrt{n}
\end{aligned}\)
\(\begin{aligned} T&=\dfrac1{2+\sqrt5}+\dfrac1{\sqrt5+\sqrt6}+\cdots+\dfrac1{\sqrt{2022}+\sqrt{2023}}\\[4pt] &=\sqrt5-2+\sqrt6-\sqrt5+\cdots+\sqrt{2023}-\sqrt{2022}\\ &=\sqrt{2023}-2\\ &=44,...-2\\ &=42,...\\ \left\lfloor T\right\rfloor&=\color{blue}\boxed{\boxed{\color{black}42}} \end{aligned}\)
\(\begin{aligned} T&=\dfrac1{2+\sqrt5}+\dfrac1{\sqrt5+\sqrt6}+\cdots+\dfrac1{\sqrt{2022}+\sqrt{2023}}\\[4pt] &=\sqrt5-2+\sqrt6-\sqrt5+\cdots+\sqrt{2023}-\sqrt{2022}\\ &=\sqrt{2023}-2\\ &=44,...-2\\ &=42,...\\ \left\lfloor T\right\rfloor&=\color{blue}\boxed{\boxed{\color{black}42}} \end{aligned}\)
Jadi, ⌊T⌋ = 42.
No.
Sederhanakan $\dfrac{2\times3}{1\times4}+\dfrac{5\times6}{4\times7}+\dfrac{8\times9}{7\times10}+\cdots+\dfrac{98\times99}{97\times100}$ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{(n+1)(n+2)}{n(n+3)}&=\dfrac{n^2+3n+2}{n^2+3n}\\[4pt]
&=1+\dfrac2{n(n+3)}\\[4pt]
&=1+\dfrac23\left(\dfrac1n-\dfrac1{n+3}\right)
\end{aligned}\)
\(\begin{aligned} \dfrac{2\times3}{1\times4}+\dfrac{5\times6}{4\times7}+\dfrac{8\times9}{7\times10}+\cdots+\dfrac{98\times99}{97\times100}&=\left(\dfrac{97-1}{4-1}+1\right)+\dfrac23\left(\dfrac11-\dfrac14+\dfrac14-\dfrac17+\cdots+\dfrac1{99}-\dfrac1{100}\right)\\[4pt] &=33+\dfrac23\left(\dfrac11-\dfrac1{100}\right)\\[4pt] &=33+\dfrac23\left(\dfrac{99}{100}\right)\\[4pt] &=33+\dfrac{33}{50}\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{1683}{50}}} \end{aligned}\)
\(\begin{aligned} \dfrac{2\times3}{1\times4}+\dfrac{5\times6}{4\times7}+\dfrac{8\times9}{7\times10}+\cdots+\dfrac{98\times99}{97\times100}&=\left(\dfrac{97-1}{4-1}+1\right)+\dfrac23\left(\dfrac11-\dfrac14+\dfrac14-\dfrac17+\cdots+\dfrac1{99}-\dfrac1{100}\right)\\[4pt] &=33+\dfrac23\left(\dfrac11-\dfrac1{100}\right)\\[4pt] &=33+\dfrac23\left(\dfrac{99}{100}\right)\\[4pt] &=33+\dfrac{33}{50}\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{1683}{50}}} \end{aligned}\)
Jadi, $\dfrac{2\times3}{1\times4}+\dfrac{5\times6}{4\times7}+\dfrac{8\times9}{7\times10}+\cdots+\dfrac{98\times99}{97\times100}=\dfrac{1683}{50}$.
No.
Diberikan $x=101!\left(\dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\cdots+\dfrac{99}{100!}\right)$. Tentukan nilai dariALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{n}{(n+1)!}&=\dfrac{n+1-1}{(n+1)!}\\[4pt]
&=\dfrac{n+1}{(n+1)!}-\dfrac1{(n+1)!}\\[4pt]
&=\dfrac{n+1}{(n+1)\cdot n!}-\dfrac1{(n+1)!}\\[4pt]
&=\dfrac1{n!}-\dfrac1{(n+1)!}
\end{aligned}\)
\(\begin{aligned} \dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\cdots+\dfrac{99}{100!}&=\dfrac1{1!}-\dfrac1{2!}+\dfrac1{2!}-\dfrac1{3!}+\dfrac1{3!}-\dfrac1{4!}+\cdots+\dfrac1{99!}-\dfrac1{100!}\\[4pt] &=\dfrac1{1!}-\dfrac1{100!}\\[4pt] &=1-\dfrac1{100!} \end{aligned}\)
\(\begin{aligned} x&=101!\left(1-\dfrac1{100!}\right)\\[4pt] &=101!-101 \end{aligned}\)
\(\begin{aligned} 101!-x&=101!-(101!-101)\\ &=\color{blue}\boxed{\boxed{\color{black}101}} \end{aligned}\)
\(\begin{aligned} \dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\cdots+\dfrac{99}{100!}&=\dfrac1{1!}-\dfrac1{2!}+\dfrac1{2!}-\dfrac1{3!}+\dfrac1{3!}-\dfrac1{4!}+\cdots+\dfrac1{99!}-\dfrac1{100!}\\[4pt] &=\dfrac1{1!}-\dfrac1{100!}\\[4pt] &=1-\dfrac1{100!} \end{aligned}\)
\(\begin{aligned} x&=101!\left(1-\dfrac1{100!}\right)\\[4pt] &=101!-101 \end{aligned}\)
\(\begin{aligned} 101!-x&=101!-(101!-101)\\ &=\color{blue}\boxed{\boxed{\color{black}101}} \end{aligned}\)
Jadi, 101! − x = 101 .
No.
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 dariALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{n}{(n+1)!}&=\dfrac{n+1-1}{(n+1)!}\\[4pt]
&=\dfrac{n+1}{(n+1)!}-\dfrac1{(n+1)!}\\[4pt]
&=\dfrac{n+1}{(n+1)\cdot n!}-\dfrac1{(n+1)!}\\[4pt]
&=\dfrac1{n!}-\dfrac1{(n+1)!}
\end{aligned}\)
\(\begin{aligned} \dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\cdots+\dfrac{2024}{2025!}&=\dfrac1{1!}-\dfrac1{2!}+\dfrac1{2!}-\dfrac1{3!}+\dfrac1{3!}-\dfrac1{4!}+\cdots+\dfrac1{2024!}-\dfrac1{2025!}\\[4pt] &=\dfrac1{1!}-\dfrac1{2025!}\\[4pt] &=1-\dfrac1{2025!} \end{aligned}\)
1 + 2025 = 2026
\(\begin{aligned} \dfrac1{2!}+\dfrac2{3!}+\dfrac3{4!}+\cdots+\dfrac{2024}{2025!}&=\dfrac1{1!}-\dfrac1{2!}+\dfrac1{2!}-\dfrac1{3!}+\dfrac1{3!}-\dfrac1{4!}+\cdots+\dfrac1{2024!}-\dfrac1{2025!}\\[4pt] &=\dfrac1{1!}-\dfrac1{2025!}\\[4pt] &=1-\dfrac1{2025!} \end{aligned}\)
1 + 2025 = 2026
Jadi, a + b = 2026.
No.
Diketahui \(S=1+\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\cdots+\dfrac1{2^{10}}\). Tentukan nilai dari \(\dfrac{S}{2-S}\).ALTERNATIF PENYELESAIAN
\(\begin{aligned}
S&=1+\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\cdots+\dfrac1{2^{10}}\\[4pt]
&=1+\dfrac12+\dfrac14+\dfrac18+\cdots+\dfrac1{1024}\\[4pt]
&=\dfrac{1023}{1024}
\end{aligned}\)
\(\begin{aligned} \dfrac{S}{2-S}&=\dfrac{\dfrac{1023}{1024}}{2-\dfrac{1023}{1024}}\\[4pt] &=\dfrac{\dfrac{1023}{1024}}{\dfrac{2048-1023}{1024}}\\[4pt] &=\dfrac{1023}{1025}\\[4pt] &=\dfrac{99}{100} \end{aligned}\)
\(\begin{aligned} \dfrac{S}{2-S}&=\dfrac{\dfrac{1023}{1024}}{2-\dfrac{1023}{1024}}\\[4pt] &=\dfrac{\dfrac{1023}{1024}}{\dfrac{2048-1023}{1024}}\\[4pt] &=\dfrac{1023}{1025}\\[4pt] &=\dfrac{99}{100} \end{aligned}\)
Jadi, \(\dfrac{S}{2-S}=\dfrac{99}{100}\).
No.
Hitunglah nilai dari $$\frac1{1\sqrt2+2\sqrt1}+\frac1{2\sqrt3+3\sqrt2}+\cdots+\frac1{\left(2014^2-1\right)\sqrt{2014^2}+2014^2\sqrt{2014^2-1}}$$ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac1{n\sqrt{n+1}+(n+1)\sqrt{n}}&=\dfrac1{\left(\sqrt{n}\right)^2\sqrt{n+1}+\left(\sqrt{n+1}\right)^2\sqrt{n}}\\[4pt]
&=\dfrac1{\sqrt{n}\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}{\color{red}\cdot\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}}\\[4pt]
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}(n+1-n)}\\[4pt]
&=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\\[4pt]
&=\dfrac1{\sqrt{n}}-\dfrac1{\sqrt{n+1}}
\end{aligned}\)
\(\begin{aligned} \dfrac1{1\sqrt2+2\sqrt1}+\dfrac1{2\sqrt3+3\sqrt2}+\cdots+\dfrac1{\left(2014^2-1\right)\sqrt{2014^2}+2014^2\sqrt{2014^2-1}}&=\left(\dfrac1{\sqrt1}-\dfrac1{\sqrt2}\right)+\left(\dfrac1{\sqrt2}-\dfrac1{\sqrt3}\right)+\cdots+\left(\dfrac1{\sqrt{2014^2-1}}-\dfrac1{\sqrt{2014^2}}\right)\\[4pt] &=1-\dfrac1{\sqrt{2014^2}}\\[4pt] &=1-\dfrac1{2014}\\[4pt] &=\dfrac{2013}{2014} \end{aligned}\)
\(\begin{aligned} \dfrac1{1\sqrt2+2\sqrt1}+\dfrac1{2\sqrt3+3\sqrt2}+\cdots+\dfrac1{\left(2014^2-1\right)\sqrt{2014^2}+2014^2\sqrt{2014^2-1}}&=\left(\dfrac1{\sqrt1}-\dfrac1{\sqrt2}\right)+\left(\dfrac1{\sqrt2}-\dfrac1{\sqrt3}\right)+\cdots+\left(\dfrac1{\sqrt{2014^2-1}}-\dfrac1{\sqrt{2014^2}}\right)\\[4pt] &=1-\dfrac1{\sqrt{2014^2}}\\[4pt] &=1-\dfrac1{2014}\\[4pt] &=\dfrac{2013}{2014} \end{aligned}\)
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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