HOTS Zone : Barisan dan Deret

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

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StandarSNBTHOTS

12

No. 1

Tentukan nilai dari
S = 1 + 2⋅2 + 3⋅2² + 4⋅2³ + ⋯ + 2022⋅2²⁰²¹.

1. 2²⁰²²
2. 2021⋅2²⁰²² − 1
3. 2021⋅2²⁰²²

4. 2021⋅2²⁰²² + 1
5. 2²⁰²² − 1

Matematika Zone Id

ALTERNATIF PENYELESAIAN

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

Jadi, S = 1 + 2⋅2 + 3⋅2² + 4⋅2³ + ⋯ + 2022⋅2²⁰²¹ = 2021⋅2²⁰²² + 1.
JAWAB: D

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No. 2

Nilai dari \left(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{\frac46} adalah

1. \dfrac46
2. \dfrac49

3. \dfrac4{11}
4. \dfrac4{13}

Grup Tricky MN

ALTERNATIF PENYELESAIAN

k⋅2k⋅4k = 8k³
k⋅3k⋅9k = 27k³ $$\begin{array}{rl}{\left({\displaystyle \frac{1\cdot 2\cdot 4+2\cdot 4\cdot 8+\cdots +n\cdot 2n\cdot 4n}{1\cdot 3\cdot 9+2\cdot 6\cdot 18+\cdots +n\cdot 3n\cdot 9n}}\right)}^{\frac{4}{6}}& ={\left({\displaystyle \frac{8\cdot 1^3+8\cdot 2^3+\cdots +8n^3}{27\cdot 1^3+27\cdot 2^3+\cdots +27n^3}}\right)}^{\frac{2}{3}}\\ & ={\left({\displaystyle \frac{8(1^3+2^3+\cdots +n^3)}{27(1^3+2^3+\cdots +n^3)}}\right)}^{\frac{2}{3}}\\ & ={\left({\displaystyle \frac{8}{27}}\right)}^{\frac{2}{3}}\\ & ={\left({\displaystyle \frac{2^3}{3^3}}\right)}^{\frac{2}{3}}\\ & ={\displaystyle \frac{2^2}{3^2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{9}}}}}}\end{array}$$

Jadi, \left(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{\frac46}=\dfrac49.
JAWAB: B

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No. 3

Carilah nilai \dfrac{3^2+1}{3^2-1}+\dfrac{5^2+1}{5^2-1}+\dfrac{7^2+1}{7^2-1}+\dots+\dfrac{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 $$\begin{array}{rl}{\displaystyle \frac{3^2+1}{3^2-1}}+{\displaystyle \frac{5^2+1}{5^2-1}}+{\displaystyle \frac{7^2+1}{7^2-1}}+\cdots +{\displaystyle \frac{{99}^2+1}{{99}^2-1}}& =1\cdot 49+{\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}+\cdots +{\displaystyle \frac{1}{48}}-{\displaystyle \frac{1}{49}}+{\displaystyle \frac{1}{49}}-{\displaystyle \frac{1}{50}})\\ & =49+{\displaystyle \frac{1}{2}}(1-{\displaystyle \frac{1}{50}})\\ & =49+{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{49}{50}}\right)\\ & =49+{\displaystyle \frac{49}{100}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4949}{100}}}}}}\end{array}$$

Jadi, \dfrac{3^2+1}{3^2-1}+\dfrac{5^2+1}{5^2-1}+\dfrac{7^2+1}{7^2-1}+\dots+\dfrac{99^2+1}{99^2-1}=\dfrac{4949}{100}.

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No. 4

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

1. \dfrac1{2!}+\dfrac1{2022!}
2. \dfrac1{2!}+\dfrac1{2021!}
3. \dfrac1{2!}-\dfrac1{2021!}

4. \dfrac1{2!}-\dfrac1{2022!}
5. \dfrac1{2021!}-\dfrac1{2022!}

Matematika Zone Id

ALTERNATIF PENYELESAIAN

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

Jadi, \dfrac{2+3^2}{1!+2!+3!+4!}+\dfrac{3+4^2}{2!+3!+4!+5!}+\cdots+\dfrac{2020+2021^2}{2019!+2020!+2021!+2022!}=\dfrac1{2!}-\dfrac1{2021!}.
JAWAB: C

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No. 5

Suatu hari Fara mellhat pola aneh di papan tulis kelasnya. Dia memperhatikannya terus dan mencoba mengerjakannya. Soalnya yaitu \left(\frac12\right)+2\left(\frac12\right)^2+3\left(\frac12\right)^3+4\left(\frac12\right)^4+5\left(\frac12\right)^5+\cdots Ternyata 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. 6

\dfrac3{1!+2!+3!}+\dfrac4{2!+3!+4!}+\cdots+\dfrac{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, \dfrac3{1!+2!+3!}+\dfrac4{2!+3!+4!}+\cdots+\dfrac{2021}{2019!+2020!+2021!}=\dfrac12-\dfrac1{2021!}.

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No. 7

Bilangan-bilangan bulat positif a_k, k = 1, 2, ⋯, 8 memenuhi persamaan \displaystyle\sum_{k=1}^8\left(k\times a_k\right)^2=204. Hasil dari \displaystyle\sum_{k=1}^8a_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 ≤ k ≤ 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}^8a_k=8.

Related: loading

JAWAB: C

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No. 8

Tentukan bentuk umum barisan yang didefinisikan oleh x₀ = 3, x₁ = 4 dan

x_{n + 1} = x_{n − 1}² − nx_n

untuk semua n ∈ 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 ≤ 1, jika x_k − 1 = k + 2 dan x_k = k + 3 maka
x_k + 1 = x_k − 1² − kx_k = (k + 2)² − k(k + 3) = k + 4

Jadi, x_n = n + 3.

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No. 9

Diberikan barisan U_n = (1, −1, 1, −1, ⋯) dengan n bilangan asli. Semua yang berikut merupakan rumus untuk barisan itu, kecuali....

1. U_n=\sin\left(n-\dfrac12\right)\pi
2. U_n = cos (n − 1)π

3. U_n = sin (n − 1)π
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)π.
JAWAB: C

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No. 10

Perhatikan gambar berikut.

Banyaknya bulatan hitam pada gambar kesepuluh nantinya adalah ....

OSK SMP 2003

ALTERNATIF PENYELESAIAN

Banyak bulatan hitam pada gambar ke-n adalah (4 + (n − 1)²). Banyak bulatan hitam pada gambar ke-10 adalah
4 + 9² = 4 + 81 = 85

Jadi, banyaknya bulatan hitam pada gambar kesepuluh nantinya adalah 85.

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