TURUNAN FUNGSI

 Berikut Kumpulan Materi Arsip Lama

• materi turunan fungsi aljabar
• bentuk umum turunan aljabar
• sifat turunan pertama dan aturan rantai
• aturan rantai pada turunan pertama dan kedua
• penggunaan turunan fungsi
• menyelesaikan masalah terkait keekstriman fungsi
• contoh soal 1, contoh 2, contoh 3, contoh 4

• materi turunan fungsi trigonometri
• lanjutan materi 
• contoh soal  (bagaian 1)
• lanjutan materi 2 
• lanjutan materi 3
• lanjutan materi 4 (sifat-sifat turunan)
• lanjutan materi 5
• contoh soal 1 (bagain 2) , contoh 2 , contoh 3 contoh 4 , contoh 5
• lanjutan materi 6 (persamaan garis singgung kurva)-turunan pertama
• lanjutan materi 7 (fungsi naik dan fungsi turun)-turunan pertama
• lanjutan materi 8 (nilai stasioner)-turunan pertama
• lanjutan materi 9 (aplikasi nilai stasioner)-turunan pertama
• contoh 6 , contoh 7 , contoh 8 , contoh 9
• turunan kedua fungsi trigonometri (Fungsi naik dan fungsi turun, titik belok serta selang kecekungan)
• contoh 10 , contoh 11 , contoh 12, contoh 13.

CONTOH SOAL 9 VISUALISASI BARISAN DAN DERET

$\begin{array}{ll}\\ 17.\quad&\textrm{Perhatikan hubungan berikut}\\ &\qquad\qquad\qquad\begin{aligned}1=&1^{\displaystyle 2}\\ 2+3+4=&3^{\displaystyle 2}\\ 3+4+5+6+7=&5^{2}\\ 4+5+6+7+8+9+10=&7^{\displaystyle 2}\\ \vdots\qquad &\vdots \end{aligned}\\ &\textrm{Tuliskan pola yang ada pada hubungan di atas dan buktikan} \end{array}$.

Bukti: $\begin{aligned}&\textrm{Pola Umum}\\ &\:\:\textrm{Baris ke}-n\:(n=1,2,3,4,...)\\ &\circ \quad \textrm{ruas kanan berupa}:(2n-1)^{\displaystyle 2}\\ &\circ \quad \textrm{ruas kiri berupa penjumlahan dari}\:\:(2n-1)\:\: \textrm{bilangan bulat berurutan}\\ &\qquad\textrm{yang dimulai dari}\:\:n\\ &\qquad n+(n+1)+(n+2)+\cdots +(3n-2)=(2n-1)^{\displaystyle 2}\\ &\qquad \text{Catatan}:U_{\displaystyle n}=n+\: (2n-1-1).1=3n-2\\ &\qquad \textrm{Bentuk umum}:\sum_{_{\displaystyle k=0}}^{^{\displaystyle 2n-2}}(n+k)=(2n-1)^{\displaystyle 2}\\ &\qquad \bullet \quad k=0\longrightarrow U_{\displaystyle 1}=n+0=n\\ &\qquad \bullet \quad k=1\longrightarrow U_{\displaystyle 2}=n+1\\ &\qquad \bullet \quad k=2\longrightarrow U_{\displaystyle 3}=n+2\\ &\qquad \bullet \quad k=3\longrightarrow U_{\displaystyle 4}=n+3\\ &\qquad\qquad \vdots \\&\qquad \bullet \quad k=2n-2\longrightarrow U_{\displaystyle n}=n+(2n-2)=3n-2\\ &\textrm{Pembuktian Pola}\\ &\quad \circ \quad \textrm{Suku pertama}=a=n\\ &\quad \circ \quad \textrm{Beda}=b=1\\ &\quad \circ \quad \textrm{Banyak suku}=2n-1\\ &\quad \circ \quad \textrm{Suku terakhir}=3n-2\\ &\quad \circ \quad \textrm{Rumus Jumlahderet aritmetika }=\displaystyle \frac{2n-1}{2}\left( a+U_{\displaystyle n} \right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\:\:\: =\displaystyle \frac{2n-1}{2}\left( n+3n-2 \right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\:\:\: =(2n-1)^{\displaystyle 2}\\\end{aligned}$.

.

Visualisasi Geometris $(2n-1)^2$

Nilai n: 3

3 + 4 + 5 + 6 + 7 = 5² = 25

LATIHAN CBT SUMATIF BARISAN DAN DERET

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CBT MA FUTUHIYAH JEKETRO

Persiapan Sumatif MA Futuhiyah Jeketro

Sumatif Matematika Kelas X

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• Alokasi Waktu: 90 Menit

• Cakupan: Pola Bilangan, Barisan & Deret Aritmetika/Geometri, Anuitas & Aplikasi.

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LATIHAN SOAL GOOGLE FORMULIR BARISAN DAN DERET

CONTOH SOAL 8 BARISAN DAN DERET

 

$\begin{aligned}15.\quad &\textrm{Di antara bilangan}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\:\: \textrm{terdapat tak hingga banyak}\\ &\textrm{bilangan pecah. Tentukan 999 bilangan pecah di antara}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\\&\textrm{sehingga selisih antara bilangan pecah berikutnya dengan bilangan}\\ &\textrm{pecah sebelumnya konstan}\\ &(\textrm{Maksudnya: jika}\:\: x_{\displaystyle 1},x_{\displaystyle 2},\cdots ,x_{\displaystyle 999}\:\:\textrm{bilangan pecah yang dimaksud}\\ &\textrm{maka},\:\:x_{\displaystyle 2}-x_{\displaystyle 1}=x_{\displaystyle 3}-x_{\displaystyle 2}=\cdots =x_{\displaystyle 999}-x_{\displaystyle 998}) \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Untuk menyisipkan 999 bilangan pecahan di antara}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\:\: \textrm{dengan}\\ &\textrm{selisih tetap, maka kita sama saja membentuk sebuah barisan aritmetika}\\&\textrm{dengan} \\ &\circ \quad \text{Suku pertama}:\:\: a=U_{\displaystyle 1}=\displaystyle \frac{1}{5}\\ &\circ \quad \text{Banyak sisipan}:\:\:999\:\: \:\textrm{bilangan pecahan}\\ &\circ \quad \text{Total suku}:k=n+2=999+2=1001\\ &\circ \quad \text{Suku terakhir}:\:\: U_{\displaystyle 1001}=\displaystyle \frac{1}{4}\\ &\circ \quad \text{Beda antar dua suku}:\:\: U_{\displaystyle n}=a+(n-1)b=\displaystyle \frac{1}{4}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow \displaystyle \frac{1}{5}+(1001-1)b=\displaystyle \frac{1}{4}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow 1000b=\displaystyle \frac{1}{4}-\frac{1}{5}=\displaystyle \frac{1}{20}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow b=\displaystyle \frac{1}{20.000}\\ &\textrm{Sehingga bilangan pecahan yang dimaksud adalah}:\\&\circ \quad U_{\displaystyle 1}=\displaystyle \frac{1}{5}=\frac{4000}{20000}\\ &\circ \quad U_{\displaystyle 2}=U_{\displaystyle 1}+b=x_{\displaystyle 1}=\frac{4000}{20000}+\frac{1}{20000}=\displaystyle \frac{4001}{20000}\\ &\circ \quad U_{\displaystyle 3}=U_{\displaystyle 2}+b=x_{\displaystyle 2}=\frac{4001}{20000}+\frac{1}{20000}=\displaystyle \frac{4002}{20000}\\ &\qquad\qquad\qquad\vdots\\ &\circ \quad U_{\displaystyle 1000}=U_{\displaystyle 999}+b=x_{\displaystyle 999}=\frac{4998}{20000}+\frac{1}{20000}=\displaystyle \frac{4999}{20000} \end{aligned}$

$\begin{aligned}16.\quad&(\textbf{LM UGM ke-32 Th 2021 Tk.SMA})\\&\textrm{Diberikan barisan}\:(a_{\displaystyle n})\: \textrm{yang memenuhi}\\ &\qquad\qquad\qquad a_{\displaystyle n+1}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{\sqrt{4(a_{\displaystyle n-1})^{\displaystyle 2}-4(a_{\displaystyle n})^{\displaystyle 2}}}\\ &\textrm{Jika}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{\sqrt{2}}\:\: \textrm{dan}\:\: a_{\displaystyle 2}=\displaystyle \frac{1}{2},\: \textrm{nilai dari}\:\: 2^{\displaystyle 2526}\prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}\:\: \textrm{adalah}\:....\\ &\text{a}.\quad \displaystyle \frac{1}{4}\qquad \qquad \qquad\qquad\qquad\:\: \text{d}.\quad \displaystyle 2\\ &\text{b}.\quad \displaystyle \frac{1}{2}\qquad\qquad \text{c}.\quad 1\qquad\qquad \text{e}.\quad 4\end{aligned}$

Jawaban: $\begin{aligned}&\textbf{Jawab: d}\\&\textrm{Perhatikan bahwa}\\ &a_{\displaystyle n+1}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{\sqrt{4(a_{\displaystyle n-1})^{\displaystyle 2}-4(a_{\displaystyle n})^{\displaystyle 2}}}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{2\sqrt{a_{\displaystyle n-1}^{\displaystyle 2}-a_{\displaystyle n}^{\displaystyle 2}}}\\ &\Leftrightarrow a_{\displaystyle n+1}^{\displaystyle 2}=\displaystyle \frac{a_{\displaystyle n}^{\displaystyle 2}a_{\displaystyle n-1}^{\displaystyle 2}}{\displaystyle 4\left( a_{\displaystyle n-1}^{\displaystyle 2}-a_{\displaystyle n}^{\displaystyle 2} \right)}\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle n+1}^{\displaystyle 2}}=\displaystyle \frac{4}{a_{\displaystyle n}^{\displaystyle 2}}-\frac{4}{a_{\displaystyle n-1}^{\displaystyle 2}}\\ &\textrm{dengan}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{\sqrt{2}}\:,\: a_{\displaystyle 2}=\displaystyle \frac{1}{2},\: \textrm{maka}\:\: a_{\displaystyle 1}^{\displaystyle 2}=\displaystyle \frac{1}{2}\:\: \textrm{dan}\:\: a_{\displaystyle 2}^{\displaystyle 2}=\displaystyle \frac{1}{4}\\ &\textrm{Misalkan saja}\:\: b_{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}},\: \textrm{maka}\:\: b_{\displaystyle n+1}=4\left( b_{\displaystyle n}-b_{\displaystyle n-1} \right)\\ &\textrm{dengan}\:\: b_{\displaystyle 1}=2,\: b_{\displaystyle 2}=4,\: \textrm{maka akan diperoleh}:\\ &\circ \quad b_{\displaystyle 3}=4(4-2)=4.2=8\\ &\circ \quad b_{\displaystyle 4}=4(8-4)=4.4=16\\ &\textrm{Tampak bahwa}\:\: b_{\displaystyle n}=2^{\displaystyle n}\: \textrm{dan}\: b_{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}},\\ &\textrm{maka}\:\: 2^{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}}\Leftrightarrow a_{\displaystyle n}=\displaystyle 2^{\displaystyle -\frac{n}{2}}.\\ &\textrm{Karena}\:\: a_{\displaystyle n}\gt 0,\:\: \textrm{maka}\:\: \prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}=\prod_{\displaystyle n=1}^{\displaystyle 100}2^{\displaystyle -\frac{1}{2}n}\\ &=2^{\displaystyle -\frac{1}{2}(1+2+3+4+\cdots +99+100)}=2^{\displaystyle -\frac{1}{2}\left( \frac{100\times 101}{2} \right)}\\ &=2^{\displaystyle -2525}\\ &\textrm{Akibatnya adalah}:\\ &2^{\displaystyle 2526}\prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}=2^{\displaystyle 2526}\times 2^{\displaystyle -2525}=2^{\displaystyle 1}=2 \end{aligned}$

CONTOH SOAL 7 BARISAN DAN DERET

 

$\begin{array}{ll}\\ 13.&\textrm{Syarat untuk deret geometri tak hingga }\\ &\textrm{dengan suku pertama}\: \: a\: \: \textrm{konvergen dengan }\\ &\textrm{jumlah 2 adalah}\: ....\:.\\ &\textrm{a}.\quad -2< a< 0\\ &\textrm{b}.\quad -4< a< 0\\ &\textrm{c}.\quad 0< a< 2\\ &\textrm{d}.\quad 0< a< 4\\ &\textrm{e}.\quad -4< a< 4\\ \end{array}$

Jawaban: $\begin{aligned}&\textbf{Jawab}:\\&\textrm{Diketahui bahwa}\: \: S_{\infty }=2,\: \: \textrm{dengan}\\ &S_{\infty }=\displaystyle \frac{a}{1-r}\Leftrightarrow 1-r=\displaystyle \frac{a}{S_{\infty }}\Leftrightarrow r=1-\displaystyle \frac{a}{S_{\infty }}\\ &\Leftrightarrow -1< 1-\displaystyle \frac{a}{S_{\infty }}< 1\Leftrightarrow -2< -\displaystyle \frac{a}{S_{\infty }}< 0\\ &\Leftrightarrow 0< \displaystyle \frac{a}{S_{\infty }}< 2\Leftrightarrow \Leftrightarrow 0< \displaystyle \frac{a}{2}< 2\\ & \Leftrightarrow 0< a<4 \end{aligned}$

$\begin{aligned}14.\quad &\textrm{Didefinisikan}\quad t_{\displaystyle n}=\displaystyle \frac{t_{\displaystyle n-1}-1}{t_{\displaystyle n-1}+1}\quad \textrm{untuk}\quad n\ge n\quad \textrm{dan}\quad t_{\displaystyle 1}=2\\ &\textrm{Berapakah nilai}\quad t_{\displaystyle 2026}? \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Untuk menentukan}\quad t_{\displaystyle 2026}\quad \textrm{kita dapat menghitung beberapa suku}\\ &\textrm{pertama dari}\quad t_{\displaystyle n}\quad \textrm{untuk menentukan pola keberulangannya, yaitu}:\\ &\textrm{Diketahui}\quad t_{\displaystyle 1}=2\quad \textrm{dan relasi rekursif}\quad t_{\displaystyle n}=\displaystyle \frac{t_{\displaystyle n-1}-1}{t_{\displaystyle n-1}+1}\\ &\circ \quad \text{Suku kedua}\:(n=2)\Rightarrow t_{\displaystyle 2}=\displaystyle \frac{t_{1}-1}{t_{1}+1}=\displaystyle \frac{2-1}{2+1}=\frac{1}{3}\\ &\circ \quad \text{Suku ketiga}\:(n=3)\Rightarrow t_{\displaystyle 3}=\displaystyle \frac{t_{2}-1}{t_{2}+1}=\displaystyle \frac{\left( \displaystyle \frac{1}{3} \right)-1}{\left( \displaystyle \frac{1}{3} \right)+1}=-\frac{1}{2}\\ &\circ \quad \text{Suku keempat}\:(n=4)\Rightarrow t_{\displaystyle 4}=\displaystyle \frac{t_{3}-1}{t_{3}+1}=\displaystyle \frac{\left( \displaystyle -\frac{1}{2} \right)-1}{\left( \displaystyle -\frac{1}{2} \right)+1}=-3\\ &\circ \quad \text{Suku kelima}\:(n=5)\Rightarrow t_{\displaystyle 5}=\displaystyle \frac{t_{4}-1}{t_{4}+1}=\displaystyle \frac{-3-1}{-3+1}=2\\ &\textrm{Karena}\quad t_{\displaystyle 5}=t_{\displaystyle 1}=2,\quad \textrm{maka nilai-nilai suku akan berulang setiap}\\ &\textrm{4 periode dengan}:\\ &\qquad\qquad\qquad t_{\displaystyle n}=\begin{cases} 2&\textrm{untuk}\quad n\equiv 1\:(\textrm{mod 4}) \\\displaystyle \frac{1}{3}&\textrm{untuk}\quad n\equiv 2\:(\textrm{mod 4})\\ \displaystyle -\frac{1}{2}&\textrm{untuk}\quad n\equiv 3\:(\textrm{mod 4})\\ -3&\textrm{untuk}\quad n\equiv 0\:(\textrm{mod 4}) \end{cases}\\ &\textrm{Untuk}\quad n=2026=4\times506+2\Rightarrow 2026\equiv 2\: (\textrm{mod 4})\\ &\textrm{Jadi, nilai}\quad t_{\displaystyle 2026}=t_{\displaystyle 2}=\displaystyle \frac{1}{3} \end{aligned}$

CONTOH SOAL 6 BARISAN DAN DERET

 

$\begin{aligned}11.\quad &\textrm{Diberikan}\\ &\quad A=1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\cdots\\ &\textrm{dan}\\ &\quad B=1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\displaystyle \frac{1}{9^{\displaystyle 4}}+\cdots\\ &\textrm{Nyatakan}\quad \displaystyle \frac{A}{B}\quad \textrm{sebagai pecahan} \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Perhatikan bahwa}\\ &A=1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{6^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots\\ &\textrm{Selanjutnya deret di ataskita bagi menjadi dua bagian, yaitu}:\\&A=\underset{\textrm{suku-suku genap}}{\underbrace{\left( \displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{6^{\displaystyle 4}}+\displaystyle \frac{1}{8^{\displaystyle 4}}+\cdots \right)}}+\underset{\textrm{suku-suku ganjil}}{\underbrace{\left( 1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots \right)}}\\ &A=\displaystyle \frac{1}{2^{\displaystyle 4}}\left( 1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\cdots \right)+\left( 1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots \right)\\ &A=\displaystyle \frac{1}{2^{\displaystyle 4}}A+B\Leftrightarrow A=\displaystyle \frac{1}{16}A+B\Leftrightarrow \displaystyle \frac{15}{16}A=B\Leftrightarrow \displaystyle \frac{A}{B}=\displaystyle \frac{16}{15}\\ &\textrm{Jadi, perbandingan}\quad \displaystyle \frac{A}{B}=\displaystyle \frac{16}{15} \end{aligned}$

$\begin{array}{ll}\\ 12.&\textbf{UM UGM}\\ &\textrm{Jumlah deret geometri tak hingga adalah 6}\\ & \textrm{Jika tiap suku dikuadratkan, maka jumlahnya}\\ &\textrm{adalah}\: \: 4\: .\: \textrm{Suku pertama deret ini adalah}\: ....\\ &\textrm{a}.\quad \displaystyle \frac{2}{5}\: \: \qquad\qquad\qquad\qquad\quad\:   \textrm{d}.\quad \displaystyle \frac{5}{6}\\ &\textrm{b}.\quad \displaystyle \frac{3}{5}\qquad\qquad \textrm{c}.\quad \displaystyle \frac{4}{5}\qquad\quad \textrm{e}.\quad \displaystyle \frac{6}{5}\\ \end{array}$

Jawaban: $\begin{aligned}&\textbf{Jawab}:\\ &\textrm{DG}=\textrm{Deret Geometri}\\ &a+ar+ar^{2}+\cdots =S_{\infty }=\displaystyle \frac{a}{1-r}=6\\ &\Leftrightarrow a=6(1-r)=6-6r\: ............(1)\\ &\textrm{Saat dikuadratkan masing-masing sukunya}\\ &a^{2}+a^{2}r^{2}+a^{2}r^{4}+\cdots =S_{\infty }=\displaystyle \frac{a^{2}}{1-r^{2}}=4\\ &\Leftrightarrow a^{2}=4(1-r^{2})=4-4r^{2}\: .......(2)\\ &\textrm{Substitusi (1) ke (2), maka} \\ &a^{2}=a^{2}\\ &\Leftrightarrow (6-6r)^{2}=4-4r^{2}\\ &\Leftrightarrow 36-72r+36r^{2}=4-4r^{2}\\ &\Leftrightarrow 40r^{2}-72r+32=0\\ &\Leftrightarrow (5r-4)(r-1)=0\\ &\Leftrightarrow r=\displaystyle \frac{4}{5}\: (memenuhi)\: \: \textbf{atau}\: \: r=1\: (tidak)\\ &\textrm{Selanjutnya kita tentukan nilai}\: \: a,\\ &a=6-6\left ( \displaystyle \frac{4}{5} \right )=6\left ( \displaystyle \frac{1}{5} \right )=\displaystyle \frac{6}{5} \end{aligned}$

CONTOH SOAL 5 BARISAN DAN DERET

 

$\begin{aligned}9.\quad&\textrm{Bilangan}\quad X=\underset{50}{\underbrace{99+999+9999+\cdots +999\cdots 999}}\:. \: \textrm{Berapa banyak}\\ &\textrm{digit 1 muncul pada bilangan}\quad X\\ & \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Suku-suku dalam dalam deret dapat dinyatakan dengan}:10^{\displaystyle k}-1\\ &\circ \quad99=10^{\displaystyle 2}-1\\ &\circ \quad 999=10^{\displaystyle 3}-1\\ &\circ \quad 9999=10^{\displaystyle 4}-1\\ &\circ \qquad \vdots\\ &\circ \quad \textrm{Suku ke}-50\: \textrm{dengan 51 digit 9}=10^{\displaystyle 51}-1\\ &\textrm{Maka persamaan}\quad X\quad \textrm{dapat dituliskan sebagai}:\\ &X=\left( 10^{\displaystyle 2}-1 \right)+\left( 10^{\displaystyle 3}-1 \right)+\left( 10^{\displaystyle 4}-1 \right)+\cdots +\left( 10^{\displaystyle 51}-1 \right)\\ &\,\quad =10^{\displaystyle 51}+10^{\displaystyle 50}+10^{\displaystyle 49}+\cdots +10^{\displaystyle 4} +10^{\displaystyle 3} +10^{\displaystyle 2}-(1\times 50)\\ &\,\quad =\underset{50\: \textrm{buah angka}\: 1}{\underbrace{111\cdots 111}00}-(1\times 50)\\ &\,\quad =\underset{49\: \textrm{buah angka}\: 1}{\underbrace{111\cdots 111}050}\\ &\textrm{Jadi, banyak digit 1 pada bilangan}\:\: X\:\: \textrm{sebanyak 49} \end{aligned}$

$\begin{aligned}10.\quad&\text{Diberikan}\:\: S=\displaystyle \frac{1+2}{2}+\displaystyle \frac{1+2+3}{2^{\displaystyle 2}}+\displaystyle \frac{1+2+3+4}{2^{\displaystyle 3}}+\cdots \\ &\textrm{Tentukan nilai}\:\: S\end{aligned}$

Jawaban: $\begin{aligned}&\text{Diketahui}\\ &S=\displaystyle \frac{1+2}{2}+\displaystyle \frac{1+2+3}{2^{\displaystyle 2}}+\displaystyle \frac{1+2+3+4}{2^{\displaystyle 3}}+\displaystyle \frac{1+2+3+4+5}{2^{\displaystyle 4}}\cdots \\ &\Leftrightarrow S=\displaystyle \frac{3}{2}+\displaystyle \frac{6}{2^{\displaystyle 2}}+\displaystyle \frac{10}{2^{\displaystyle 3}}+\displaystyle \frac{15}{2^{\displaystyle 4}}+\displaystyle \frac{21}{2^{\displaystyle 5}}+\displaystyle \frac{28}{2^{\displaystyle 6}}+\cdots \\ &\Leftrightarrow \displaystyle \frac{1}{2}S=\displaystyle \frac{3}{2^{\displaystyle 2}}+\displaystyle \frac{6}{2^{\displaystyle 3}}+\displaystyle \frac{10}{2^{\displaystyle 4}}+\displaystyle \frac{15}{2^{\displaystyle 5}}+\displaystyle \frac{21}{2^{\displaystyle 6}}+\cdots \\ &\text{Untuk}\quad S-\displaystyle \frac{1}{2}S=\displaystyle \frac{3}{2}+\displaystyle \frac{6-3}{2^{\displaystyle 2}}+\displaystyle \frac{10-6}{2^{\displaystyle 3}}+\displaystyle \frac{15-10}{2^{\displaystyle 4}}+\displaystyle \frac{21-15}{2^{\displaystyle 5}}+\cdots \\ &\,\:\qquad\qquad \Leftrightarrow \displaystyle \frac{1}{2}S=\displaystyle \frac{3}{2}+\displaystyle \frac{3}{2^{\displaystyle 2}}+\displaystyle \frac{4}{2^{\displaystyle 3}}+\displaystyle \frac{5}{2^{\displaystyle 4}}+\displaystyle \frac{6}{2^{\displaystyle 5}}+\cdots \\ &\textrm{Saat}\quad \displaystyle \frac{1}{2}\left( \displaystyle \frac{1}{2}S \right)=\displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2^{\displaystyle 2}}+\displaystyle \frac{3}{2^{\displaystyle 3}}+\displaystyle \frac{4}{2^{\displaystyle 4}}+\displaystyle \frac{5}{2^{\displaystyle 5}}+\cdots \\ &\textrm{dan}\: \left( \displaystyle \frac{1}{2}S-\displaystyle \frac{1}{4}S \right)=\displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2}+0+\displaystyle \frac{1}{2^{\displaystyle 3}}+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{2^{\displaystyle 5}}+\cdots \\ &\Leftrightarrow \displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2}+\underset{\textrm{deret geometri tak hingga}}{\underbrace{\displaystyle \frac{\displaystyle \frac{1}{2^{\displaystyle 3}}}{1-\displaystyle \frac{1}{2}}}}\\ &\Leftrightarrow \displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2}+\displaystyle \frac{1}{4}\qquad (\textrm{masing-masing ruas dikali 4})\\ &\Leftrightarrow S=6+1=7\\ &\textrm{Jadi, nilai}\quad S=7\end{aligned}$

CONTOH SOAL 4 BARISAN DAN DERET

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

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}\\$.

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

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}$

CONTOH SOAL 3 BARISAN DAN DERET

$\begin{aligned}5.\quad&(\textbf{KSM Matematika MA Tk. Kab/kota Th.2013})\\ &\textrm{Hasil penjumlahan dari tak hingga suku berbentuk}\\ &\displaystyle \frac{1}{5}+\frac{2}{5^{\displaystyle 2}}+\frac{1}{5^{\displaystyle 3}}+\frac{2}{5^{\displaystyle 4}}+\frac{1}{5^{\displaystyle 5}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\quad \textrm{adalah}\: ....\\\\ &\text{a}.\quad \displaystyle \frac{25}{24}\qquad\qquad\qquad\text{d}.\quad \displaystyle \frac{1}{4}\\ &\text{b}.\quad \displaystyle \frac{24}{25}\qquad\qquad\quad\quad\text{e}.\quad \displaystyle \frac{1}{12}\\ &\text{c}.\quad \displaystyle \frac{7}{24}\\\end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Perhatikan bahwa deret jumlah semua sukunya}:\\ &\displaystyle \frac{1}{5}+\frac{2}{5^{\displaystyle 2}}+\frac{1}{5^{\displaystyle 3}}+\frac{2}{5^{\displaystyle 4}}+\frac{1}{5^{\displaystyle 5}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\\ &\textrm{Selanjutnya deret kita partisi menjadi dua bagian, yaitu}:\\ &\textbf{Deret pertama untuk suku ganjil}\\ &S_{\displaystyle 1}=\displaystyle \frac{1}{5}+\frac{1}{5^{\displaystyle 3}}+\frac{1}{5^{\displaystyle 5}}\:+\:...\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=a=\displaystyle \frac{1}{5} \\\textrm{rasio}&=r=\displaystyle \frac{1}{25}\\ S_{\displaystyle \infty }&=\displaystyle \frac{a}{1-r}=\frac{\displaystyle \frac{1}{5}}{1-\displaystyle \frac{1}{25}}=\displaystyle \frac{5}{24} \end{cases}\\ &\textbf{Deret kedua untuk suku genap}\\ &S_{\displaystyle 2}=\frac{2}{5^{\displaystyle 2}}+\frac{2}{5^{\displaystyle 4}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=a=\displaystyle \frac{2}{5^{\displaystyle 2}}=\frac{2}{25} \\\textrm{rasio}&=r=\displaystyle \frac{1}{25}\\ S_{\displaystyle \infty }&=\displaystyle \frac{a}{1-r}=\frac{\displaystyle \frac{2}{25}}{1-\displaystyle \frac{1}{25}}=\displaystyle \frac{2}{24} \end{cases}\\ &\textbf{Jumlah totalnya adalah}:\\ &S_{\textrm{total}}=S_{\displaystyle 1}+S_{\displaystyle 2}=\displaystyle \frac{5}{24}+\frac{2}{24}=\displaystyle \frac{7}{24}\end{aligned}$

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$\begin{aligned}6.\quad&(\textbf{KSM Matematika MA Tk. Kab/kota Th.2013})\\ &\textrm{nilai}\quad n\quad \textrm{terkecil yang memenuhi}\:\: \frac{1}{2^{\displaystyle n}}\lt 0,001\:\: \textrm{adalah}\: ....\\ &\text{a}.\quad \displaystyle 9\qquad\qquad\qquad\text{d}.\quad \displaystyle 522\\ &\text{b}.\quad \displaystyle 10\:\:\quad\quad\quad\quad\quad\text{e}.\quad \displaystyle 501\\ &\text{c}.\quad \displaystyle 11\\ \end{aligned}$

Jawaban: $\begin{aligned}&\textbf{Jawab: b}\\ &\textrm{Perhatikan bahwa}:\\ &\frac{1}{2^{\displaystyle n}}\lt 0,001\Leftrightarrow \displaystyle \frac{1}{2^{\displaystyle n}}\lt \displaystyle \frac{1}{1000}\Leftrightarrow 2^{\displaystyle n}\gt 1000\\ &\textrm{Selanjutnya cukup kita uji untuk}\\&\begin{cases} n=9&\Rightarrow 2^{\displaystyle 9}=512\ngtr 1000 \\n=10&\Rightarrow 2^{\displaystyle 10}=1024\gt 1000\\ n=11&\Rightarrow 2^{\displaystyle 11}=2048\gt 1000\: (\textrm{bukan yang terkecil}) \end{cases}\\ &\textbf{Jadi,}\:\: n\:\: \textrm{terkecilnya adalah}=10\end{aligned}$