CONTOH SOAL 4 BARISAN

$\begin{aligned}7.\quad&\textrm{Perhatikanlah hubungan berikut}\\ &\qquad\qquad\begin{matrix} 6^{\displaystyle 2}-5^{\displaystyle 2}=11 \\ 56^{\displaystyle 2}-45^{\displaystyle 2}=1111 \\ 556^{\displaystyle 2}-445^{\displaystyle 2}=111111 \\ 5556^{\displaystyle 2}-4445^{\displaystyle 2}=11111111 \\ \vdots \quad\qquad\qquad \vdots \end{matrix}\\ &\textrm{Tulislah pola yang ada pada hubungan di atas dan buktikan}\\& \end{aligned}$.

Bukti: $\begin{aligned} &\textrm{Perhatikan bahwa}:\\ &\textbf{Pola suku pertama}:\:\textrm{ angka 6 yang didahului oleh}\quad n\quad \textrm{buah}\\ &\textrm{angka 5. Polanya dapat dituliskan sebagai}:\\ &A_{\displaystyle n}=\underset{\textrm{n buah 5}}{\underbrace{55\cdots 56}}=\displaystyle \frac{5.\left( 10^{\displaystyle n+1}-1 \right)}{9}+1=\displaystyle \frac{5.10^{\displaystyle n+1}+4}{9}\\ &\textbf{Pola suku kedua}:\:\textrm{ angka 5 yang didahului oleh}\quad n\quad \textrm{buah}\\ &\textrm{angka 4. Polanya dapat dituliskan sebagai}:\\ &B_{\displaystyle n}=\underset{\textrm{n buah 4}}{\underbrace{44\cdots 45}}=\displaystyle \frac{4.\left( 10^{\displaystyle n+1}-1 \right)}{9}+1=\displaystyle \frac{4.10^{\displaystyle n+1}+5}{9}\\ &\textbf{Hasil(ruas kanan)}\\ &\underset{\textrm{2n+2 buah 1}}{\underbrace{11\cdots 11}}=\displaystyle \frac{10^{\displaystyle 2n+2}-1}{9}\\ &\textbf{Pola umum}\\ &\left( \underset{\textrm{n }}{\underbrace{55\cdots 56}} \right)^{\displaystyle 2}-\left( \underset{\textrm{n}}{\underbrace{44\cdots 45}} \right)^{\displaystyle 2}=\underset{\textrm{2n+2}}{\underbrace{11\cdots 11}}\\ &\textrm{Ingat bentuk selisih kuadrat, yaitu}:a^{2}-b^{2}=(a+b)(a-b)\\ &A_{\displaystyle n}^{\displaystyle 2}-B_{\displaystyle n}^{\displaystyle 2}=\left( A_{\displaystyle n}+B_{\displaystyle n} \right)\left( A_{\displaystyle n}-B_{\displaystyle n} \right)\\ &=\left( \displaystyle \frac{5.10^{\displaystyle n+1}+4}{9}+\displaystyle \frac{4.10^{\displaystyle n+1}+5}{9} \right)\left( \displaystyle \frac{5.10^{\displaystyle n+1}+4}{9}-\displaystyle \frac{4.10^{\displaystyle n+1}+5}{9} \right)\\&=\left( \frac{9.10^{\displaystyle n+1}+9}{9} \right)\left( \displaystyle \frac{10^{\displaystyle n+1}-1}{9} \right)\\ &=\left( 10^{\displaystyle n+1}+1 \right)\left( \displaystyle \frac{10^{\displaystyle n+1}-1}{9} \right)\\ &=\displaystyle \frac{\left( 10^{\displaystyle n+1} \right)^{\displaystyle 2}-1}{9}\\ &=\displaystyle \frac{\left( 10^{\displaystyle 2n+2} \right)-1}{9}\quad \textrm{adalah deretan angka 1 sebanyak}\quad 2n+2\\ &=\underset{2n+2}{\underbrace{11\cdots 11}}\qquad \textbf{Terbukti}\end{aligned}\\$.

.

$\begin{aligned}8.\quad&\textrm{Barisan}\:\: \left\{ a_{\displaystyle n} \right\}_{n\ge 1}\:\: \textrm{didefinisikan dengan}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{2},\: a_{\displaystyle k+1}=a_{\displaystyle k}^{\displaystyle 2}+a_{\displaystyle k}\\ &\textrm{untuk semua}\:\: k\ge 1.\:\: \textrm{Tentukan bilangan bulat terbesar yang kurang}\\ &\textrm{dari atau sama dengan}\: \displaystyle \frac{1}{a_{\displaystyle 1}+1}+\frac{1}{a_{\displaystyle 2}+1}+\cdots +\frac{1}{a_{\displaystyle 2026}+1}\\\end{aligned}$

[ Lihat Jawaban ]

$\begin{aligned}&\textrm{Diberikan barisan}:a_{\displaystyle 1},a_{\displaystyle 2},a_{\displaystyle 3},\cdots \:\: \textrm{dengan}\: a_{\displaystyle 1}=\displaystyle \frac{1}{2},a_{\displaystyle k+1}=a_{\displaystyle k}^{\displaystyle 2}+a_{\displaystyle k}\\ &a_{\displaystyle k+1}=a_{\displaystyle k}\left( a_{\displaystyle k}+1 \right)\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle k+1}}=\displaystyle \frac{1}{a_{\displaystyle k}\left( a_{\displaystyle k}+1 \right)}\\ &\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle k+1}}=\displaystyle \frac{1}{a_{\displaystyle k}}-\displaystyle \frac{1}{a_{\displaystyle k}+1}\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle k}+1}=\displaystyle \frac{1}{a_{\displaystyle k}}-\displaystyle \frac{1}{a_{\displaystyle k+1}}\\ &\textrm{Selanjutnya kembali ke deret pada soal}\\ &S_{\displaystyle n}=\displaystyle \sum_{k=1}^{n}\displaystyle \frac{1}{a_{\displaystyle k}+1}=\left( \displaystyle \frac{1}{a_{\displaystyle 1}}-\frac{1}{a_{\displaystyle 2}} \right)+\left( \displaystyle \frac{1}{a_{\displaystyle 2}}-\frac{1}{a_{\displaystyle 3}} \right)+\cdots +\left( \displaystyle \frac{1}{a_{\displaystyle n}}-\frac{1}{a_{\displaystyle n+1}} \right)\\ &\:\:\:\,\quad\quad\quad\quad\quad\quad\quad=\displaystyle \frac{1}{a_{\displaystyle 1}}-\displaystyle \frac{1}{a_{\displaystyle n+1}}=\displaystyle \frac{1}{\left( \displaystyle \frac{1}{2} \right)}-\displaystyle \frac{1}{a_{\displaystyle n+1}}=2-\displaystyle \frac{1}{a_{\displaystyle n+1}}\\ &S_{\displaystyle 2026}=\displaystyle \sum_{k=1}^{n}\displaystyle \frac{1}{a_{\displaystyle 2026}+1}=2-\displaystyle \frac{1}{a_{\displaystyle 2027}}\\ &\qquad\qquad\qquad\qquad\qquad\qquad(\textrm{dengan}\quad a_{\displaystyle 2027}\gt 1\Rightarrow 1\lt \displaystyle \frac{1}{a_{\displaystyle 2027}}\lt 2)\\ &\textrm{Jadi, bilangan bulat terbesar yang kurang dari atau sama dengan}\\& S_{\displaystyle 2026}=\left\lfloor 2-\displaystyle \frac{1}{a_{\displaystyle 2027}} \right\rfloor =1 \end{aligned}$