HOTS Zone : Barisan dan Deret

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Tipe:

Standar SBMPTN HOTS

No.

Tentukan nilai dari $S=1+2\cdot 2+3\cdot 2^2+4\cdot 2^3+\cdots +2022\cdot 2^{2021}$.

1. $2^{2022}$
2. $2021\cdot 2^{2022}-1$
3. $2021\cdot 2^{2022}$

4. $2021\cdot 2^{2022}+1$
5. $2^{2022}-1$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}S& =1+2\cdot 2+3\cdot 2^2+4\cdot 2^3+\cdots +2022\cdot 2^{2021}\\ 2S& =1\cdot 2+2\cdot 2^2+3\cdot 2^3+\cdots +2021\cdot 2^{2021}+2022\cdot 2^{2022}-\\ -S& =1+2+2^2+2^3+\cdots +2^{2021}-2022\cdot 2^{2022}\\ & ={\displaystyle \frac{1(2^{2022}-1)}{2-1}}-2022\cdot 2^{2022}\\ & =2^{2022}-1-2022\cdot 2^{2022}\\ & =-2021\cdot 2^{2022}-1\\ S& =\boxed{{\displaystyle \boxed{{\displaystyle 2021\cdot 2^{2022}+1}}}}\end{array}$

Jadi, $S=1+2\cdot 2+3\cdot 2^2+4\cdot 2^3+\cdots +2022\cdot 2^{2021}=2021\cdot 2^{2022}+1$.
JAWAB: D

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No.

Nilai dari ${\left({\displaystyle \frac{1\cdot 2\cdot 4+2\cdot 4\cdot 8+\cdots +n\cdot 2n\cdot 4n}{1\cdot 3\cdot 9+2\cdot 6\cdot 18+\cdots +n\cdot 3n\cdot 9n}}\right)}^{\frac{4}{6}}$ adalah

1. ${\displaystyle \frac{4}{6}}$
2. ${\displaystyle \frac{4}{9}}$

3. ${\displaystyle \frac{4}{11}}$
4. ${\displaystyle \frac{4}{13}}$

SUMBER

ALTERNATIF PENYELESAIAN

$k\cdot 2k\cdot 4k=8k^3$
$k\cdot 3k\cdot 9k=27k^3$

$\begin{array}{rl}{\left({\displaystyle \frac{1\cdot 2\cdot 4+2\cdot 4\cdot 8+\cdots +n\cdot 2n\cdot 4n}{1\cdot 3\cdot 9+2\cdot 6\cdot 18+\cdots +n\cdot 3n\cdot 9n}}\right)}^{\frac{4}{6}}& ={\left({\displaystyle \frac{8\cdot 1^3+8\cdot 2^3+\cdots +8n^3}{27\cdot 1^3+27\cdot 2^3+\cdots +27n^3}}\right)}^{\frac{2}{3}}\\ & ={\left({\displaystyle \frac{8(1^3+2^3+\cdots +n^3)}{27(1^3+2^3+\cdots +n^3)}}\right)}^{\frac{2}{3}}\\ & ={\left({\displaystyle \frac{8}{27}}\right)}^{\frac{2}{3}}\\ & ={\left({\displaystyle \frac{2^3}{3^3}}\right)}^{\frac{2}{3}}\\ & ={\displaystyle \frac{2^2}{3^2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{9}}}}}}\end{array}$

Jadi, ${\left({\displaystyle \frac{1\cdot 2\cdot 4+2\cdot 4\cdot 8+\cdots +n\cdot 2n\cdot 4n}{1\cdot 3\cdot 9+2\cdot 6\cdot 18+\cdots +n\cdot 3n\cdot 9n}}\right)}^{\frac{4}{6}}={\displaystyle \frac{4}{9}}$.
JAWAB: B

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No.

Carilah nilai ${\displaystyle \frac{3^2+1}{3^2-1}}+{\displaystyle \frac{5^2+1}{5^2-1}}+{\displaystyle \frac{7^2+1}{7^2-1}}+\cdots +{\displaystyle \frac{{99}^2+1}{{99}^2-1}}$

SUMBER

ALTERNATIF PENYELESAIAN

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

$99=2(49)+1$

$\begin{array}{rl}{\displaystyle \frac{3^2+1}{3^2-1}}+{\displaystyle \frac{5^2+1}{5^2-1}}+{\displaystyle \frac{7^2+1}{7^2-1}}+\cdots +{\displaystyle \frac{{99}^2+1}{{99}^2-1}}& =1+{\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}+\cdots +{\displaystyle \frac{1}{48}}-{\displaystyle \frac{1}{49}}+{\displaystyle \frac{1}{49}}-{\displaystyle \frac{1}{50}})\\ & =1+{\displaystyle \frac{1}{2}}(1-{\displaystyle \frac{1}{50}})\\ & =1+{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{49}{50}}\right)\\ & =1+{\displaystyle \frac{49}{100}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{149}{100}}}}}}\end{array}$

Jadi, ${\displaystyle \frac{3^2+1}{3^2-1}}+{\displaystyle \frac{5^2+1}{5^2-1}}+{\displaystyle \frac{7^2+1}{7^2-1}}+\cdots +{\displaystyle \frac{{99}^2+1}{{99}^2-1}}={\displaystyle \frac{149}{100}}$.

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No.

Nilai dari ekspresi:
${\displaystyle \frac{2+3^2}{1!+2!+3!+4!}}+{\displaystyle \frac{3+4^2}{2!+3!+4!+5!}}+\cdots +{\displaystyle \frac{2020+{2021}^2}{2019!+2020!+2021!+2022!}}$ adalah....

1. ${\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{2022!}}$
2. ${\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{2021!}}$
3. ${\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{2021!}}$

4. ${\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{2022!}}$
5. ${\displaystyle \frac{1}{2021!}}-{\displaystyle \frac{1}{2022!}}$

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{(k+1)+(k+2{)}^2}{k!+(k+1)!+(k+2)!+(k+3)!}}& ={\displaystyle \frac{k+1+k^2+4k+4}{k!(1+(k+1)+(k+2)(k+1)+(k+3)(k+2)(k+1))}}\\ & ={\displaystyle \frac{k+1+k^2+4k+4}{k!((k+2)+(k+2)(k+1)+(k+2)(k^2+4k+3))}}\\ & ={\displaystyle \frac{k+1+k^2+4k+4}{k!(k+2)(1+k+1+k^2+4k+3)}}\\ & ={\displaystyle \frac{k^2+5k+5}{k!(k+2)(k^2+5k+5)}}\\ & ={\displaystyle \frac{1}{k!(k+2)}}\\ & ={\displaystyle \frac{k+1}{(k+2)!}}\\ & ={\displaystyle \frac{1}{(k+1)!}}-{\displaystyle \frac{1}{(k+2)!}}\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{2+3^2}{1!+2!+3!+4!}}+{\displaystyle \frac{3+4^2}{2!+3!+4!+5!}}+\cdots +{\displaystyle \frac{2020+{2021}^2}{2019!+2020!+2021!+2022!}}& ={\displaystyle \frac{1}{2!}}-\cancel{{\displaystyle \frac{1}{3!}}}+\cancel{{\displaystyle \frac{1}{3!}}}-\cancel{{\displaystyle \frac{1}{4!}}}+\cdots +\cancel{{\displaystyle \frac{1}{2020!}}}-{\displaystyle \frac{1}{2021!}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{2021!}}}}}}\end{array}$$

Jadi, ${\displaystyle \frac{2+3^2}{1!+2!+3!+4!}}+{\displaystyle \frac{3+4^2}{2!+3!+4!+5!}}+\cdots +{\displaystyle \frac{2020+{2021}^2}{2019!+2020!+2021!+2022!}}={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{2021!}}$.
JAWAB: C

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No.

Suatu hari Fara mellhat pola aneh di papan tulis kelasnya. Dia memperhatikannya terus dan mencoba mengerjakannya. Soalnya yaitu $$\left(\frac{1}{2}\right)+2{\left(\frac{1}{2}\right)}^2+3{\left(\frac{1}{2}\right)}^3+4{\left(\frac{1}{2}\right)}^4+5{\left(\frac{1}{2}\right)}^5+\cdots$$ Ternyara soal tersebut menggunakan konsep deret dalam pengerjaannya. Langkah yang dilakukan oleh Fara yaitu dengan memisalkan bahwa deret yang di atas sama dengan $x$. Maka berapakah nilai dari $x$?

Facebook

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}x& =\left(\frac{1}{2}\right)+2{\left(\frac{1}{2}\right)}^2+3{\left(\frac{1}{2}\right)}^3+4{\left(\frac{1}{2}\right)}^4+5{\left(\frac{1}{2}\right)}^5+\cdots \\ {\displaystyle \frac{1}{2}}x& ={\left(\frac{1}{2}\right)}^2+2{\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^4+4{\left(\frac{1}{2}\right)}^5+\cdots & -\\ {\displaystyle \frac{1}{2}}x& =\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^5+\cdots \\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{2}}}}\\ & =1\\ x& =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, $x=2$.

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No.

${\displaystyle \frac{3}{1!+2!+3!}}+{\displaystyle \frac{4}{2!+3!+4!}}+\cdots +{\displaystyle \frac{2021}{2019!+2020!+2021!}}=$ ....

SUMBER (dengan sedikit perubahan)

ALTERNATIF PENYELESAIAN

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

$\begin{array}{rl}{\displaystyle \frac{3}{1!+2!+3!}}+{\displaystyle \frac{4}{2!+3!+4!}}+\cdots +{\displaystyle \frac{2021}{2019!+2020!+2021!}}& ={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{3!}}-{\displaystyle \frac{1}{4!}}+\cdots +{\displaystyle \frac{1}{2020!}}-{\displaystyle \frac{1}{2021!}}\\ & ={\displaystyle \frac{1}{2!}}-{\displaystyle \frac{1}{2021!}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2021!}}}}}}\end{array}$

Jadi, ${\displaystyle \frac{3}{1!+2!+3!}}+{\displaystyle \frac{4}{2!+3!+4!}}+\cdots +{\displaystyle \frac{2021}{2019!+2020!+2021!}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2021!}}$.

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No.

Bilangan-bilangon bulat positif $a_k$, $k=1,2,K,8$ memenuhi persamaan ${\displaystyle \sum _{k=1}^8{(k\times a_k)}^2=204}$. Hasil dari ${\displaystyle \sum _{k=1}^8{a}_k}$

1. $6$
2. $7$
3. $8$

4. $10$
5. $12$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \sum _{k=1}^n{k}^2}& ={\displaystyle \frac{n(n+1)(2n+1)}{6}}\\ {\displaystyle \sum _{k=1}^8{k}^2}& ={\displaystyle \frac{8(8+1)(2\cdot 8+1)}{6}}\\ & =204\end{array}$
Sehingga $a_k=1$ untuk $1\le k\le 8$

$\begin{array}{rl}{\displaystyle \sum _{k=1}^8{a}_k}& =1\cdot 8\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, ${\displaystyle \sum _{k=1}^8{a}_k=8}$.
JAWAB: C

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No.

Tentukan bentuk umum barisan yang didefinisikan oleh $x_0=3$, $x_1=4$ dan

$x_{n+1}=x_{n-1}^2-n{x}_n$

untuk semua $n\in N$.

101 Problems in Algebra

ALTERNATIF PENYELESAIAN

Kita bisa buktikan dengan induksi bahwa $x_n=n+3$. Mudah dibuktikan benar untuk $n=0$ dan $n=1$.
Untuk $k\le 1$, jika $x_{k-1}=k+2$ dan $x_k=k+3$ maka
$x_{k+1}=x_{k-1}^2-k{x}_k=(k+2{)}^2-k(k+3)=k+4$

Jadi, $x_n=n+3$.

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No.

Diberikan barisan $U_n=(1,-1,1,-1,\cdots )$ dengan $n$ bilangan asli. Semua yang berikut merupakan rumus untuk barisan itu, kecuali....

1. $U_n=\sin (n-{\displaystyle \frac{1}{2}})\pi$
2. $U_n=\cos (n-1)\pi$

3. $U_n=\sin (n-1)\pi$
4. $U_n=\{\begin{array}{l}1,\text{jika n ganjil}\\ -1,\text{jika n genap}\end{array}$

Grup PPG Matematika

ALTERNATIF PENYELESAIAN

Substitusikan $n=1$. Hanya opsi C yang tidak menghasilkan 1.

Jadi, kecuali $U_n=\sin (n-1)\pi$.
JAWAB: C

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