HOTS Zone : Deret Teleskopik

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STANDARSNBTHOTS

No. 1

Tentukan nilai dari ${\displaystyle \frac{3}{1!+2!+3!}}+{\displaystyle \frac{4}{2!+3!+4!}}+{\displaystyle \frac{5}{3!+4!+5!}}+\cdots +{\displaystyle \frac{1634}{1632!+1633!+1634!}}$

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{k+2}{k!+(k+1)!+(k+2)!}}& ={\displaystyle \frac{k+2}{k!(1+(k+1)+(k+2)(k+1))}}\\ & ={\displaystyle \frac{k+2}{k!(1+k+1+k^2+3k+2)}}\\ & ={\displaystyle \frac{k+2}{k!(k^2+4k+4)}}\\ & ={\displaystyle \frac{k+2}{k!{(k+2)}^2}}\\ & ={\displaystyle \frac{1}{k!(k+2)}}\\ & ={\displaystyle \frac{k+1}{(k+2)!}}\\ & ={\displaystyle \frac{k+2-1}{(k+2)!}}\\ & ={\displaystyle \frac{1}{(k+1)!}}-{\displaystyle \frac{1}{(k+2)!}}\end{array}$

$\begin{array}{rl}{\displaystyle \frac{3}{1!+2!+3!}}+{\displaystyle \frac{4}{2!+3!+4!}}+{\displaystyle \frac{5}{3!+4!+5!}}+\cdots +{\displaystyle \frac{1634}{1632!+1633!+1634!}}& ={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{3!}}-{\displaystyle \frac{1}{4!}}+{\displaystyle \frac{1}{4!}}-{\displaystyle \frac{1}{5!}}+\cdots +{\displaystyle \frac{1}{1633!}}-{\displaystyle \frac{1}{1634!}}\\ & ={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{1634!}}\end{array}$

Jadi, ${\displaystyle \frac{3}{1!+2!+3!}}+{\displaystyle \frac{4}{2!+3!+4!}}+{\displaystyle \frac{5}{3!+4!+5!}}+\cdots +{\displaystyle \frac{1634}{1632!+1633!+1634!}}={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{1634!}}$.

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No. 2

Hitunglah nilai ${\displaystyle \sum _{k=1}^{1634}{\displaystyle \frac{k+2}{k(k+1)2^k}}}$

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \sum _{k=1}^{1634}{\displaystyle \frac{k+2}{k(k+1)2^k}}}& ={\displaystyle \sum _{k=1}^{1634}({\displaystyle \frac{2}{k}}-{\displaystyle \frac{1}{k+1}}){\displaystyle \frac{1}{2^k}}}\\ & ={\displaystyle \sum _{k=1}^{1634}({\displaystyle \frac{1}{k\cdot 2^{k-1}}}-{\displaystyle \frac{1}{(k+1)\cdot 2^k}})}\\ & ={\displaystyle \frac{1}{1\cdot 2^0}}-{\displaystyle \frac{1}{2\cdot 2^1}}+{\displaystyle \frac{1}{2\cdot 2^1}}-{\displaystyle \frac{1}{3\cdot 2^2}}+\cdots +{\displaystyle \frac{1}{1634\cdot 2^{1633}}}-{\displaystyle \frac{1}{1635\cdot 2^{1634}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1-{\displaystyle \frac{1}{1635\cdot 2^{1634}}}}}}}\end{array}$

Jadi, ${\displaystyle \sum _{k=1}^{1634}{\displaystyle \frac{k+2}{k(k+1)2^k}}=1-{\displaystyle \frac{1}{1635\cdot 2^{1634}}}}$.

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No. 3

Berapa hasil dari:

$(x+1)(x+7349)({\displaystyle \frac{1}{(x+1)(x+2)}}+{\displaystyle \frac{1}{(x+2)(x+3)}}+\cdots +{\displaystyle \frac{1}{(x+7348)(x+7349)}})$

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}(x+1)(x+7349)({\displaystyle \frac{1}{(x+1)(x+2)}}+{\displaystyle \frac{1}{(x+2)(x+3)}}+\cdots +{\displaystyle \frac{1}{(x+7348)(x+7349)}})& =(x+1)(x+7349)({\displaystyle \frac{1}{x+1}}-{\displaystyle \frac{1}{x+2}}+{\displaystyle \frac{1}{x+2}}-{\displaystyle \frac{1}{x+3}}+\cdots +{\displaystyle \frac{1}{x+7348}}-{\displaystyle \frac{1}{x+7349}})\\ & =(x+1)(x+7349)({\displaystyle \frac{1}{x+1}}-{\displaystyle \frac{1}{x+7349}})\\ & =(x+7349)-(x+1)\\ & =7349-1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 7348}}}}\end{array}$$

Jadi, $(x+1)(x+7349)({\displaystyle \frac{1}{(x+1)(x+2)}}+{\displaystyle \frac{1}{(x+2)(x+3)}}+\cdots +{\displaystyle \frac{1}{(x+7348)(x+7349)}})=7348$.

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No. 4

$${\displaystyle \frac{1}{1\times 2}}+{\displaystyle \frac{1}{2\times 3}}+{\displaystyle \frac{1}{3\times 4}}+\cdots +{\displaystyle \frac{1}{118\times 119}}=....$$

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{1}{1\times 2}}+{\displaystyle \frac{1}{2\times 3}}+{\displaystyle \frac{1}{3\times 4}}+\cdots +{\displaystyle \frac{1}{118\times 119}}& ={\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}+\cdots +{\displaystyle \frac{1}{118}}-{\displaystyle \frac{1}{119}}\\ & ={\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{119}}\\ & ={\displaystyle \frac{119-1}{119}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{118}{119}}}}}}\end{array}$$

Jadi, ${\displaystyle \frac{1}{1\times 2}}+{\displaystyle \frac{1}{2\times 3}}+{\displaystyle \frac{1}{3\times 4}}+\cdots +{\displaystyle \frac{1}{118\times 119}}={\displaystyle \frac{118}{119}}$.

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