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No.
Tentukan nilai dari $S=1+2\cdot2+3\cdot2^2+4\cdot2^3+\cdots+2022\cdot2^{2021}$.- $2^{2022}$
- ${2021\cdot2^{2022}-1}$
- $2021\cdot2^{2022}$
- ${2021\cdot2^{2022}+1}$
- $2^{2022}-1$
ALTERNATIF PENYELESAIAN
$\begin{aligned}
S&=1+2\cdot2+3\cdot2^2+4\cdot2^3+\cdots+2022\cdot2^{2021}\\
2S&=1\cdot2+2\cdot2^2+3\cdot2^3+\cdots+2021\cdot2^{2021}+2022\cdot2^{2022}\qquad\color{red}-\\\hline
-S&=1+2+2^2+2^3+\cdots+2^{2021}-2022\cdot2^{2022}\\
&=\dfrac{1\left(2^{2022}-1\right)}{2-1}-2022\cdot2^{2022}\\
&=2^{2022}-1-2022\cdot2^{2022}\\
&=-2021\cdot2^{2022}-1\\
S&=\boxed{\boxed{2021\cdot2^{2022}+1}}
\end{aligned}$
No.
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- $\dfrac46$
- $\dfrac49$
- $\dfrac4{11}$
- $\dfrac4{13}$
ALTERNATIF PENYELESAIAN
$k\cdot2k\cdot4k=8k^3$
$k\cdot3k\cdot9k=27k^3$
$\begin{aligned} \left(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{\frac46}&=\left(\dfrac{8\cdot1^3+8\cdot2^3+\cdots+8n^3}{27\cdot1^3+27\cdot2^3+\cdots+27n^3}\right)^{\frac23}\\[4pt] &=\left(\dfrac{8\left(1^3+2^3+\cdots+n^3\right)}{27\left(1^3+2^3+\cdots+n^3\right)}\right)^{\frac23}\\[8pt] &=\left(\dfrac8{27}\right)^{\frac23}\\[8pt] &=\left(\dfrac{2^3}{3^3}\right)^{\frac23}\\[8pt] &=\dfrac{2^2}{3^2}\\ &=\boxed{\boxed{\dfrac49}} \end{aligned}$
$k\cdot3k\cdot9k=27k^3$
$\begin{aligned} \left(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{\frac46}&=\left(\dfrac{8\cdot1^3+8\cdot2^3+\cdots+8n^3}{27\cdot1^3+27\cdot2^3+\cdots+27n^3}\right)^{\frac23}\\[4pt] &=\left(\dfrac{8\left(1^3+2^3+\cdots+n^3\right)}{27\left(1^3+2^3+\cdots+n^3\right)}\right)^{\frac23}\\[8pt] &=\left(\dfrac8{27}\right)^{\frac23}\\[8pt] &=\left(\dfrac{2^3}{3^3}\right)^{\frac23}\\[8pt] &=\dfrac{2^2}{3^2}\\ &=\boxed{\boxed{\dfrac49}} \end{aligned}$
No.
Carilah nilai $\dfrac{3^2+1}{3^2-1}+\dfrac{5^2+1}{5^2-1}+\dfrac{7^2+1}{7^2-1}+\dots+\dfrac{99^2+1}{99^2-1}$ALTERNATIF PENYELESAIAN
$\begin{aligned}
\dfrac{(2k+1)^2+1}{(2k+1)^2-1}&=\dfrac{4k^2+4k+1+1}{4k^2+4k+1-1}\\[8pt]
&=\dfrac{4k^2+4k+2}{4k^2+4k}\\[8pt]
&=\dfrac{2k^2+2k+1}{2k^2+2k}\\[8pt]
&=1+\dfrac1{2k^2+2k}\\[8pt]
&=1+\dfrac1{2k(k+1)}\\[8pt]
&=1+\dfrac12\left(\dfrac1k-\dfrac1{k+1}\right)
\end{aligned}$
$99=2(49)+1$
$\begin{aligned} \dfrac{3^2+1}{3^2-1}+\dfrac{5^2+1}{5^2-1}+\dfrac{7^2+1}{7^2-1}+\dots+\dfrac{99^2+1}{99^2-1}&=1+\dfrac12\left(\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\cdots+\dfrac1{48}-\dfrac1{49}+\dfrac1{49}-\dfrac1{50}\right)\\[8pt] &=1+\dfrac12\left(1-\dfrac1{50}\right)\\[8pt] &=1+\dfrac12\left(\dfrac{49}{50}\right)\\[8pt] &=1+\dfrac{49}{100}\\ &=\boxed{\boxed{\dfrac{149}{100}}} \end{aligned}$
$99=2(49)+1$
$\begin{aligned} \dfrac{3^2+1}{3^2-1}+\dfrac{5^2+1}{5^2-1}+\dfrac{7^2+1}{7^2-1}+\dots+\dfrac{99^2+1}{99^2-1}&=1+\dfrac12\left(\dfrac11-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\cdots+\dfrac1{48}-\dfrac1{49}+\dfrac1{49}-\dfrac1{50}\right)\\[8pt] &=1+\dfrac12\left(1-\dfrac1{50}\right)\\[8pt] &=1+\dfrac12\left(\dfrac{49}{50}\right)\\[8pt] &=1+\dfrac{49}{100}\\ &=\boxed{\boxed{\dfrac{149}{100}}} \end{aligned}$
No.
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....
- $\dfrac1{2!}+\dfrac1{2022!}$
- $\dfrac1{2!}+\dfrac1{2021!}$
- $\dfrac1{2!}-\dfrac1{2021!}$
- $\dfrac1{2!}-\dfrac1{2022!}$
- $\dfrac1{2021!}-\dfrac1{2022!}$
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)}\\[8pt]
&=\dfrac{k+1+k^2+4k+4}{k!\left((k+2)+(k+2)(k+1)+(k+2)(k^2+4k+3)\right)}\\[8pt]
&=\dfrac{k+1+k^2+4k+4}{k!(k+2)\left(1+k+1+k^2+4k+3\right)}\\[8pt]
&=\dfrac{k^2+5k+5}{k!(k+2)\left(k^2+5k+5\right)}\\[8pt]
&=\dfrac1{k!(k+2)}\\[8pt]
&=\dfrac{k+1}{(k+2)!}\\[8pt]
&=\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}
No.
Suatu hari Fara mellhat pola aneh di papan tulis kelasnya. Dia memperhatikannya terus dan mencoba mengerjakannya. Soalnya yaitu $$\left(\frac12\right)+2\left(\frac12\right)^2+3\left(\frac12\right)^3+4\left(\frac12\right)^4+5\left(\frac12\right)^5+\cdots$$ 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 $x$?ALTERNATIF PENYELESAIAN
\(\eqalign{
x&=\left(\frac12\right)+2\left(\frac12\right)^2+3\left(\frac12\right)^3+4\left(\frac12\right)^4+5\left(\frac12\right)^5+\cdots\\
\dfrac12x&=\qquad\qquad\ \left(\frac12\right)^2+2\left(\frac12\right)^3+3\left(\frac12\right)^4+4\left(\frac12\right)^5+\cdots\qquad&{\color{red}-}\\\hline
\dfrac12x&=\left(\frac12\right)+\left(\frac12\right)^2+\left(\frac12\right)^3+\left(\frac12\right)^4+\left(\frac12\right)^5+\cdots\\
&=\dfrac{\dfrac12}{1-\dfrac12}\\
&=1\\
x&=\boxed{\boxed{2}}
}\)
No.
${\dfrac3{1!+2!+3!}+\dfrac4{2!+3!+4!}+\cdots+\dfrac{2021}{2019!+2020!+2021!}=}$ ....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} $
$\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} $
No.
Bilangan-bilangon bulat positif $a_k$, $k=1,2,K,8$ memenuhi persamaan ${\displaystyle\sum_{k=1}^8\left(k\times a_k\right)^2=204}$. Hasil dari $\displaystyle\sum_{k=1}^8a_k$- $6$
- $7$
- $8$
- $10$
- $12$
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 $a_k=1$ untuk $1\leq k\leq8$
$\begin{aligned} \displaystyle\sum_{k=1}^8a_k&=1\cdot8\\ &=\boxed{\boxed{8}} \end{aligned}$
Sehingga $a_k=1$ untuk $1\leq k\leq8$
$\begin{aligned} \displaystyle\sum_{k=1}^8a_k&=1\cdot8\\ &=\boxed{\boxed{8}} \end{aligned}$
No.
Tentukan bentuk umum barisan yang didefinisikan oleh $x_0=3$, $x_1=4$ dan$x_{n+1}=x_{n-1}^2-nx_n$
untuk semua $n\in N$.
ALTERNATIF PENYELESAIAN
Kita bisa buktikan dengan induksi bahwa $x_n=n+3$. Mudah dibuktikan benar untuk $n=0$ dan $n=1$.
Untuk $k\leq1$, jika $x_{k-1}=k+2$ dan $x_k=k+3$ maka
$x_{k+1}=x_{k-1}^2-kx_k=(k+2)^2-k(k+3)=k+4$
Untuk $k\leq1$, jika $x_{k-1}=k+2$ dan $x_k=k+3$ maka
$x_{k+1}=x_{k-1}^2-kx_k=(k+2)^2-k(k+3)=k+4$
No.
Diberikan barisan $U_n= (1, -1, 1, -1, \cdots)$ dengan $n$ bilangan asli. Semua yang berikut merupakan rumus untuk barisan itu, kecuali....- $U_n=\sin\left(n-\dfrac12\right)\pi$
- $U_n=\cos\left(n-1\right)\pi$
- $U_n=\sin\left(n-1\right)\pi$
- $U_n=\begin{cases}1,\ \text{jika n ganjil}\\-1,\ \text{jika n genap}\end{cases}$
ALTERNATIF PENYELESAIAN
Substitusikan $n=1$. Hanya opsi C yang tidak menghasilkan 1.
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