HOTS Zone : Deret Teleskopik
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 .
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