HOTS Zone : Barisan dan Deret
Table of Contents
Tipe:
No.
Tentukan nilai dariS = 1 + 2⋅2 + 3⋅22 + 4⋅23 + ⋯ + 2022⋅22021.
- 22022
- 2021⋅22022 − 1
- 2021⋅22022
- 2021⋅22022 + 1
- 22022 − 1
ALTERNATIF PENYELESAIAN
\begin{aligned}
S&=1+2\cdot2+3\cdot2^2+4\cdot2^3+\cdots+2022\cdot2^{2021}\\
2S&=1\cdot2+2\cdot2^2+3\cdot2^3+\cdots+2021\cdot2^{2021}+2022\cdot2^{2022}\qquad\color{red}-\\\hline
-S&=1+2+2^2+2^3+\cdots+2^{2021}-2022\cdot2^{2022}\\
&=\dfrac{1\left(2^{2022}-1\right)}{2-1}-2022\cdot2^{2022}\\
&=2^{2022}-1-2022\cdot2^{2022}\\
&=-2021\cdot2^{2022}-1\\
S&=\boxed{\boxed{2021\cdot2^{2022}+1}}
\end{aligned}
Jadi, S = 1 + 2⋅2 + 3⋅22 + 4⋅23 + ⋯ + 2022⋅22021 = 2021⋅22022 + 1.
JAWAB: D
JAWAB: D
No.
Nilai dari-
\dfrac46 -
\dfrac49
-
\dfrac4{11} -
\dfrac4{13}
ALTERNATIF PENYELESAIAN
k⋅2k⋅4k = 8k3
k⋅3k⋅9k = 27k3 \begin{aligned} \left(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{\frac46}&=\left(\dfrac{8\cdot1^3+8\cdot2^3+\cdots+8n^3}{27\cdot1^3+27\cdot2^3+\cdots+27n^3}\right)^{\frac23}\\[4pt] &=\left(\dfrac{8\left(1^3+2^3+\cdots+n^3\right)}{27\left(1^3+2^3+\cdots+n^3\right)}\right)^{\frac23}\\[3.5pt] &=\left(\dfrac8{27}\right)^{\frac23}\\[3.5pt] &=\left(\dfrac{2^3}{3^3}\right)^{\frac23}\\[3.5pt] &=\dfrac{2^2}{3^2}\\ &=\boxed{\boxed{\dfrac49}} \end{aligned}
k⋅3k⋅9k = 27k3 \begin{aligned} \left(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{\frac46}&=\left(\dfrac{8\cdot1^3+8\cdot2^3+\cdots+8n^3}{27\cdot1^3+27\cdot2^3+\cdots+27n^3}\right)^{\frac23}\\[4pt] &=\left(\dfrac{8\left(1^3+2^3+\cdots+n^3\right)}{27\left(1^3+2^3+\cdots+n^3\right)}\right)^{\frac23}\\[3.5pt] &=\left(\dfrac8{27}\right)^{\frac23}\\[3.5pt] &=\left(\dfrac{2^3}{3^3}\right)^{\frac23}\\[3.5pt] &=\dfrac{2^2}{3^2}\\ &=\boxed{\boxed{\dfrac49}} \end{aligned}
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
JAWAB: B
No.
Carilah nilaiALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{(2k+1)^2+1}{(2k+1)^2-1}&=\dfrac{4k^2+4k+1+1}{4k^2+4k+1-1}\\[3.5pt]
&=\dfrac{4k^2+4k+2}{4k^2+4k}\\[3.5pt]
&=\dfrac{2k^2+2k+1}{2k^2+2k}\\[3.5pt]
&=1+\dfrac1{2k^2+2k}\\[3.5pt]
&=1+\dfrac1{2k(k+1)}\\[3.5pt]
&=1+\dfrac12\left(\dfrac1k-\dfrac1{k+1}\right)
\end{aligned}
99 = 2(49) + 1
\begin{aligned}
\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}&=1\cdot49+\dfrac12\left(\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\cdots+\dfrac1{48}-\dfrac1{49}+\dfrac1{49}-\dfrac1{50}\right)\\[3.5pt]
&=49+\dfrac12\left(1-\dfrac1{50}\right)\\[3.5pt]
&=49+\dfrac12\left(\dfrac{49}{50}\right)\\[3.5pt]
&=49+\dfrac{49}{100}\\
&=\boxed{\boxed{\dfrac{4949}{100}}}
\end{aligned}
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} .
No.
Nilai dari ekspresi:-
\dfrac1{2!}+\dfrac1{2022!} -
\dfrac1{2!}+\dfrac1{2021!} -
\dfrac1{2!}-\dfrac1{2021!}
-
\dfrac1{2!}-\dfrac1{2022!} -
\dfrac1{2021!}-\dfrac1{2022!}
ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{(k+1)+(k+2)^2}{k!+(k+1)!+(k+2)!+(k+3)!}&=\dfrac{k+1+k^2+4k+4}{k!\left(1+(k+1)+(k+2)(k+1)+(k+3)(k+2)(k+1)\right)}\\[3.5pt]
&=\dfrac{k+1+k^2+4k+4}{k!\left((k+2)+(k+2)(k+1)+(k+2)(k^2+4k+3)\right)}\\[3.5pt]
&=\dfrac{k+1+k^2+4k+4}{k!(k+2)\left(1+k+1+k^2+4k+3\right)}\\[3.5pt]
&=\dfrac{k^2+5k+5}{k!(k+2)\left(k^2+5k+5\right)}\\[3.5pt]
&=\dfrac1{k!(k+2)}\\[3.5pt]
&=\dfrac{k+1}{(k+2)!}\\[3.5pt]
&=\dfrac1{(k+1)!}-\dfrac1{(k+2)!}
\end{aligned}
\begin{aligned}
\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!}-\cancel{\dfrac1{3!}}+\cancel{\dfrac1{3!}}-\cancel{\dfrac1{4!}}+\cdots+\cancel{\dfrac1{2020!}}-\dfrac1{2021!}\\
&=\boxed{\boxed{\dfrac1{2!}-\dfrac1{2021!}}}
\end{aligned}
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
JAWAB: C
No.
Suatu hari Fara mellhat pola aneh di papan tulis kelasnya. Dia memperhatikannya terus dan mencoba mengerjakannya. Soalnya yaituALTERNATIF PENYELESAIAN
\begin{aligned}
x&=\left(\frac12\right)+2\left(\frac12\right)^2+3\left(\frac12\right)^3+4\left(\frac12\right)^4+5\left(\frac12\right)^5+\cdots\\
\dfrac12x&=\qquad\qquad\ \left(\frac12\right)^2+2\left(\frac12\right)^3+3\left(\frac12\right)^4+4\left(\frac12\right)^5+\cdots&-\\\hline
\dfrac12x&=\left(\frac12\right)+\left(\frac12\right)^2+\left(\frac12\right)^3+\left(\frac12\right)^4+\left(\frac12\right)^5+\cdots\\
&=\dfrac{\dfrac12}{1-\dfrac12}\\
&=1\\
x&=\boxed{\boxed{2}}
\end{aligned}
Jadi, x = 2.
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{n+2}{n!+(n+1)!+(n+2)!}&=\dfrac{n+2}{n!+(n+1)\cdot n!+(n+2)\cdot(n+1)\cdot n!}\\
&=\dfrac{n+2}{n!\left(1+(n+1)+(n+2)\cdot(n+1)\right)}\\
&=\dfrac{n+2}{n!\left(1+n+1+n^2+3n+2\right)}\\
&=\dfrac{n+2}{n!\left(n^2+4n+4\right)}\\
&=\dfrac{n+2}{n!\left(n+2\right)^2}\\
&=\dfrac1{n!\left(n+2\right)}\\
&=\dfrac{n+1}{\left(n+2\right)\cdot(n+1)\cdot n!}\\
&=\dfrac{n+2-1}{\left(n+2\right)!}\\
&=\dfrac{n+2}{\left(n+2\right)!}-\dfrac1{\left(n+2\right)!}\\
&=\dfrac1{\left(n+1\right)!}-\dfrac1{\left(n+2\right)!}
\end{aligned}
\begin{aligned}
\dfrac3{1!+2!+3!}+\dfrac4{2!+3!+4!}+\cdots+\dfrac{2021}{2019!+2020!+2021!}&=\dfrac1{2!}-\dfrac1{3!}+\dfrac1{3!}-\dfrac1{4!}+\cdots+\dfrac1{2020!}-\dfrac1{2021!}\\
&=\dfrac1{2!}-\dfrac1{2021!}\\
&=\boxed{\boxed{\dfrac12-\dfrac1{2021!}}}
\end{aligned}
Jadi, \dfrac3{1!+2!+3!}+\dfrac4{2!+3!+4!}+\cdots+\dfrac{2021}{2019!+2020!+2021!}=\dfrac12-\dfrac1{2021!} .
No.
Bilangan-bilangan bulat positif ak, k = 1, 2, ⋯, 8 memenuhi persamaan- 6
- 7
- 8
- 10
- 12
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\sum_{k=1}^nk^2&=\dfrac{n(n+1)(2n+1)}6\\
\displaystyle\sum_{k=1}^8k^2&=\dfrac{8(8+1)(2\cdot8+1)}6\\
&=204
\end{aligned}
Sehingga ak = 1 untuk 1 ≤ k ≤ 8
\begin{aligned}
\displaystyle\sum_{k=1}^8a_k&=1\cdot8\\
&=\boxed{\boxed{8}}
\end{aligned}
Jadi, \displaystyle\sum_{k=1}^8a_k=8 .
JAWAB: C
JAWAB: C
No.
Tentukan bentuk umum barisan yang didefinisikan oleh x0 = 3, x1 = 4 danxn + 1 = xn − 12 − nxn
untuk semua n ∈ N.
ALTERNATIF PENYELESAIAN
Kita bisa buktikan dengan induksi bahwa xn = n + 3. Mudah dibuktikan benar untuk n = 0 dan n = 1.
Untuk k ≤ 1, jika xk − 1 = k + 2 dan xk = k + 3 maka
xk + 1 = xk − 12 − kxk = (k + 2)2 − k(k + 3) = k + 4
Untuk k ≤ 1, jika xk − 1 = k + 2 dan xk = k + 3 maka
xk + 1 = xk − 12 − kxk = (k + 2)2 − k(k + 3) = k + 4
Jadi, xn = n + 3.
No.
Diberikan barisan Un = (1, −1, 1, −1, ⋯) dengan n bilangan asli. Semua yang berikut merupakan rumus untuk barisan itu, kecuali....U_n=\sin\left(n-\dfrac12\right)\pi - Un = cos (n − 1)π
- Un = sin (n − 1)π
U_n=\begin{cases}1,\ \text{jika n ganjil}\\-1,\ \text{jika n genap}\end{cases}
ALTERNATIF PENYELESAIAN
Substitusikan n = 1. Hanya opsi C yang tidak menghasilkan 1.
Jadi, kecuali Un = sin (n − 1)π.
JAWAB: C
JAWAB: C
No.
Perhatikan gambar berikut. Banyaknya bulatan hitam pada gambar kesepuluh nantinya adalah ....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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