HOTS Zone : Barisan dan Deret

\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}

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}

\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}

\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}

\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}

\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} \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 ≤ k ≤ 8 \begin{aligned} \displaystyle\sum_{k=1}^8a_k&=1\cdot8\\ &=\boxed{\boxed{8}} \end{aligned}

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

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

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