Soal 01
SIMAK UI 2010 Matematika Dasar 203
\(x_1\) dan \(x_2\) adalah bilangan bulat yang merupakan akar-akar persamaan kuadrat \(x^2 - (2p + 4)x + (3p + 4) = 0\), dimana p adalah suatu konstanta. Jika \(x_1, p, x_2\) merupakan tiga suku pertama dari suatu deret geometri, maka suku ke-12 dari deret geometri tersebut adalah ...
(A) \(-1\)
(B) \(1\)
(C) \(6 + 2 \sqrt{5}\)
(D) \(6 - 2 \sqrt{5}\)
(E) \(4\)
Jawab: A
\(x^2 - (2p + 4)x + (3p + 4) = 0\)
\begin{equation*}
\begin{split}
x_1 + x_2 & = - \frac {- (2p + 4)}{1} \\\\
x_1 + x_2 & = 2p + 4
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
x_1 \:.\: x_2 & = \frac {3p + 4}{1} \\\\
x_1 \:.\: x_2 & = 3p + 4
\end{split}
\end{equation*}
Barisan geometri \(x_1, p, x_2\)
\begin{equation*}
\begin{split}
& \frac {p}{x_1} = \frac {x_2}{p} \\\\
& p^2 = x_1 \:.\: x_2 \\\\
& p^2 = 3p + 4 \\\\
& p^2 - 3p - 4 = 0 \\\\
& (p - 4)(p + 1) = 0 \\\\
& p = 4 \text{ atau } p = -1
\end{split}
\end{equation*}
Untuk \(p = 4\)
\begin{equation*}
\begin{split}
& x^2 - (2p + 4)x + (3p + 4) = 0 \\\\
& x^2 - 12x + 16 = 0 \\\\
& (x - 6)^2 - 20 = 0 \\\\
& (x - 6)^2 = 20 \\\\
& x - 6 = \pm 2 \sqrt{5} \\\\
& x = 6 \pm 2 \sqrt{5}
\end{split}
\end{equation*}
Untuk \(p = -1\)
\begin{equation*}
\begin{split}
& x^2 - (2p + 4)x + (3p + 4) = 0 \\\\
& x^2 - 2x + 1 = 0 \\\\
& (x - 1)^2 = 0 \\\\
& x - 1 = 0 \\\\
& x = 1
\end{split}
\end{equation*}
Karena \(x_1\) dan \(x_2\) adalah bilangan bulat, maka \(x_1 = 1\) dan \(x_2 = 1\)
Deret geometri adalah \(1,-1,1\)
\begin{equation*}
\begin{split}
& U_{12} = a \:.\: r^{n - 1} \\\\
& U_{12} = 1 \:.\: (-1)^{12 - 1} \\\\
& U_{12} = 1 \:.\: (-1)^{11} \\\\
& U_{12} = 1 \:.\: -1 \\\\
& \bbox[5px, border: 2px solid magenta] {U_{12} = -1}
\end{split}
\end{equation*}
Soal 02
SIMAK UI 2011 Matematika Dasar 211
\(1 - 3 + 5 + 7 - 9 + 11 + 13 - 15 + 17 + \dotso + 193 - 195 + 197 = \dotso\)
(A) 3399
(B) 3366
(C) 3333
(D) 3267
(E) 3266
Jawab: D
Menentukan jumlah \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + \dotso + 193 + 195 + 197\)
\begin{equation*}
\begin{split}
U_n & = a + (n - 1) \:.\: b \\\\
197 & = 1 + (n - 1) \:.\: 2 \\\\
196 & = (n - 1) \:.\: 2 \\\\
98 & = n - 1 \\\\
99 & = n
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
S_n & = \frac n2 \:.\: (a + U_n) \\\\
S_{99} & = \frac {99}{2} \:.\: (1 + 197) \\\\
S_{99} & = \frac {99}{2} \:.\: 198 \\\\
S_{99} & = 9801
\end{split}
\end{equation*}
Menentukan jumlah \(-3 - 9 - 15 - \dotso - 195\)
\begin{equation*}
\begin{split}
U_n & = a + (n - 1) \:.\: b \\\\
-195 & = -3 + (n - 1) \:.\: -6 \\\\
-192 & = (n - 1) \:.\: -6 \\\\
32 & = n - 1 \\\\
33 & = n
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
S_n & = \frac n2 \:.\: (a + U_n) \\\\
S_{33} & = \frac {33}{2} \:.\: (-3 - 195) \\\\
S_{33} & = \frac {33}{2} \:.\: -198 \\\\
S_{33} & = 3267
\end{split}
\end{equation*}
Maka \(9801 - 2 \:.\: (3267) = 3267\)
Soal 03
SIMAK UI 2012 Matematika Dasar 221
Diketahui sebuah barisan \(\dfrac 32, \dfrac 34, \dfrac 98, \dfrac {15}{16}, \dotso\). Jumlah sepuluh suku pertama dari barisan tersebut adalah ...
(A) \(10 + \dfrac {1 - 2^{-10}}{3}\)
(B) \(10 - \dfrac {- 2^{-10} - 1}{3}\)
(C) \(10 + \dfrac {2^{-10} - 1}{3}\)
(D) \(\dfrac {- 2^{-10} - 1}{3}\)
(E) \(10\)
Jawab: A
\begin{equation*}
\begin{split}
\frac 32 + \frac 34 + \frac 98 + \frac {15}{16} + \dotso & = m \\\\
\text{masing-masing suku dikurangi 1 sebanyak 10 suku} & \\\\
\frac 32 - {\color {blue} 1} + \frac 34 - {\color {blue} 1} + \frac 98 - {\color {blue} 1} + \frac {15}{16} - {\color {blue} 1} + \dotso & = m - {\color {blue} 10} \\\\
\frac 12 - \frac 14 + \frac 18 - \frac {1}{16} + \dotso & = m - 10 \quad {\color {red} \dotso \: (1)}
\end{split}
\end{equation*}
Bentuk \(\dfrac 12 - \dfrac 14 + \dfrac 18 - \dfrac {1}{16} + \dotso\) merupakan deret geometri dengan \(a = \dfrac 12\) dan \(r = - \dfrac 12\)
\begin{equation*}
\begin{split}
S_{10} & = \frac {a \left[1 - r^n \right]}{1 - r} \\\\
S_{10} & = \frac {\dfrac 12\left[1 - \left(- \dfrac 12 \right)^{10}\right]}{1 - \left(- \dfrac 12 \right)} \\\\
S_{10} & = \frac {\dfrac 12\left[1 - 2^{-10} \right]}{\dfrac 32} \\\\
S_{10} & = \frac 13 \left[1 - 2^{-10} \right]
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& m - 10 = \frac 12 - \frac 14 + \frac 18 - \frac {1}{16} + \dotso \quad {\color {red} \dotso \: (1)} \\\\
& m - 10 = \frac 13 \left[1 - 2^{-10} \right] \\\\
& \bbox[5px, border: 2px solid magenta] {m = 10 + \frac 13 \left[1 - 2^{-10} \right]}
\end{split}
\end{equation*}
Soal 04
SIMAK UI 2012 Matematika Dasar 221
Jika diketahui \(\sqrt{y^2 + 2y + 1}, \dfrac {y^2 + 3y - 1}{3}, y - 1\) adalah tiga suku barisan aritmetika, maka nilai suku kedua yang memenuhi adalah ...
(1) −1
(2) −2
(3) 1
(4) 2
Jawab: (1) dan (3)
Barisan aritmetika \(\sqrt{y^2 + 2y + 1}, \dfrac {y^2 + 3y - 1}{3}, y - 1\)
\begin{equation*}
\begin{split}
U_2 - U_1 & = U_3 - U_2 \\\\
\frac {y^2 + 3y - 1}{3} - \sqrt{y^2 + 2y + 1} & = y - 1 - \frac {y^2 + 3y - 1}{3} \\\\
2 \:.\: \frac {y^2 + 3y - 1}{3} + 1 - y & = \sqrt{y^2 + 2y + 1} \\\\
\frac {2y^2 + 6y - 2}{3} + \frac 33 - \frac {3y}{3} & = \sqrt{(y + 1)^2} \\\\
\frac {2y^2 + 3y + 1}{3} & = |y + 1| \\\\
\end{split}
\end{equation*}
Ada dua kemungkinan penyelesaian:
\begin{equation*}
\begin{split}
y + 1 & = \frac {2y^2 + 3y + 1}{3} \\\\
2y^2 + 3y + 1 & = 3y + 3 \\\\
2y^2 & = 2 \\\\
y^2 & = 1 \\\\
y & = \pm 1
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
y + 1 & = -\frac {2y^2 + 3y + 1}{3} \\\\
2y^2 + 3y + 1 & = -3y - 3 \\\\
2y^2 + 6y + 4 & = 0 \\\\
y^2 + 3y + 2 & = 0 \\\\
(y + 1)(y + 2) & = 0 \\\\
y = -1 \text{ atau } y & = -2
\end{split}
\end{equation*}
Nilai y yang mungkin adalah \(-2, -1, 1\)
Untuk \(y = -2\)
\begin{equation*}
\begin{split}
& U_2 = \frac {y^2 + 3y - 1}{3} \\\\
& U_2 = \frac {(-2)^2 + 3(-2) - 1}{3} \\\\
& U_2 = \frac {4 - 6 - 1}{3} \\\\
& \bbox[5px, border: 2px solid magenta] {U_2 = -1}
\end{split}
\end{equation*}
Untuk \(y = -1\)
\begin{equation*}
\begin{split}
& U_2 = \frac {y^2 + 3y - 1}{3} \\\\
& U_2 = \frac {(-1)^2 + 3(-1) - 1}{3} \\\\
& U_2 = \frac {1 - 3 - 1}{3} \\\\
& \bbox[5px, border: 2px solid magenta] {U_2 = -1}
\end{split}
\end{equation*}
Untuk \(y = 1\)
\begin{equation*}
\begin{split}
& U_2 = \frac {y^2 + 3y - 1}{3} \\\\
& U_2 = \frac {(1)^2 + 3(1) - 1}{3} \\\\
& U_2 = \frac {1 + 3 - 1}{3} \\\\
& \bbox[5px, border: 2px solid magenta] {U_2 = 1}
\end{split}
\end{equation*}
Soal 05
SIMAK UI 2013 Matematika Dasar 331
Jika diketahui bahwa
\(x = \dfrac {1}{2013} - \dfrac {2}{2013} + \dfrac {3}{2013} - \dfrac {4}{2013} + \dotso - \dfrac {2012}{2013}\), nilai x yang memenuhi adalah ...
(A) \(- \dfrac {1007}{2013}\)
(B) \(- \dfrac {1006}{2013}\)
(C) \(\dfrac {1}{2013}\)
(D) \(\dfrac {1006}{2013}\)
(E) \(\dfrac {1007}{2013}\)
Jawab: B
\begin{equation*}
\begin{split}
& x = {\color{blue}\frac {1}{2013}} - {\color{red}\frac {2}{2013}} + {\color{blue}\frac {3}{2013}} - {\color{red}\frac {4}{2013}} + \dotso + {\color{blue}\frac {2011}{2013}} - {\color{red}\frac {2012}{2013}} \\\\
& x = {\color{blue}\frac {1}{2013} + \frac {3}{2013} + \dotso + \frac {2011}{2013}} - {\color{red}\frac {2}{2013} - \frac {4}{2013} - \dotso - \frac {2012}{2013}}
\end{split}
\end{equation*}
Deret aritmetika \({\color{blue}\dfrac {1}{2013} + \dfrac {3}{2013} + \dotso + \dfrac {2011}{2013}}\)
\begin{equation*}
\begin{split}
U_n & = a + (n - 1) \:.\: b \\\\
\frac {2011}{2013} & = \frac {1}{2013} + (n - 1) \:.\: \frac {2}{2013} \\\\
\frac {2010}{2013} & = (n - 1) \:.\: \frac {2}{2013} \\\\
1005 & = n - 1 \\\\
1006 & = n
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
S_n & = \frac n2 (a + U_n) \\\\
S_{1006} & = \frac {1006}{2} \left(\frac {1}{2013} + \frac {2011}{2013} \right) \\\\
S_{1006} & = 503 \left(\frac {2012}{2013} \right)
\end{split}
\end{equation*}
Deret aritmetika \({\color{red}- \dfrac {2}{2013} - \dfrac {4}{2013} - \dotso - \dfrac {2012}{2013}}\)
\begin{equation*}
\begin{split}
U_n & = a + (n - 1) \:.\: b \\\\
- \frac {2012}{2013} & = - \frac {2}{2013} + (n - 1) \:.\: \frac {-2}{2013} \\\\
-\frac {2010}{2013} & = (n - 1) \:.\: \frac {-2}{2013} \\\\
1005 & = n - 1 \\\\
1006 & = n
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
S_n & = \frac n2 (a + U_n) \\\\
S_{1006} & = \frac {1006}{2} \left(- \frac {2}{2013} - \frac {2012}{2013} \right) \\\\
S_{1006} & = 503 \left(-\frac {2014}{2013} \right)
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& S_{\text{total}} = 503 \left(\frac {2012}{2013} \right) + 503 \left(-\frac {2014}{2013} \right) \\\\
& S_{\text{total}} = 503 \left(\frac {2012}{2013} - \frac {2014}{2013} \right) \\\\
& S_{\text{total}} = 503 \left(-\frac {2}{2013} \right) \\\\
& \bbox[5px, border: 2px solid magenta] {S_{\text{total}} = -\frac {1006}{2013}}
\end{split}
\end{equation*}
Soal 06
SIMAK UI 2013 Matematika Dasar 331
Diketahui bahwa \(x, \: a_1, \: a_2, \: a_3, \: y\) dan \(x, \: b_1, \: b_2, \: b_3, \: b_4, \: b_5, \: y\) dengan \(x \neq y\) adalah dua buah barisan aritmetika, maka \(\dfrac {a_3 - a_2}{b_5 - b_3} = \dotso\)
(A) \(\dfrac 23\)
(B) \(\dfrac 57\)
(C) \(\dfrac 34\)
(D) \(\dfrac 56\)
(E) \(\dfrac 43\)
Jawab: C
Barisan aritmetika \(x, \: a_1, \: a_2, \: a_3, \: y\)
\begin{equation*}
\begin{split}
U_5 - U_1 & = 4 \:.\: b \\\\
y - x & = 4 \:.\: b \\\\
\frac {y - x}{4} & = b
\end{split}
\end{equation*}
Barisan aritmetika \(x, \: b_1, \: b_2, \: b_3, \: b_4, \: b_5, \: y\)
\begin{equation*}
\begin{split}
U_7 - U_1 & = 6 \:.\: b' \\\\
y - x & = 6 \:.\: b' \\\\
\frac {y - x}{6} & = b'
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
\frac {a_3 - a_2}{b_5 - b_3} & = \frac {b}{2b'} \\\\
\frac {a_3 - a_2}{b_5 - b_3} & = \frac {\dfrac {y - x}{4}}{2 \:.\: \dfrac {y - x}{6}} \\\\
\frac {a_3 - a_2}{b_5 - b_3} & = \frac {\dfrac {y - x}{4}}{\dfrac {y - x}{3}} \\\\
\frac {a_3 - a_2}{b_5 - b_3} & = \frac {y - x}{4} \:.\: \frac {3}{y - x} \\\\
\frac {a_3 - a_2}{b_5 - b_3} & = \frac 34
\end{split}
\end{equation*}
Soal 07
SIMAK UI 2014 Matematika Dasar 511
Diketahui untuk \(n > 1\), berlaku \(s_n = \dfrac {1}{2^n} + \dfrac {1}{3^n} + \dfrac {1}{4^n} + \dotso\), maka \(s_2 + s_3 + s_4 + \dotso = \dotso\)
(A) \(1\)
(B) \(2\)
(C) \(\pi\)
(D) \(\pi^2\)
(E) \(\sim\)
Jawab: A
\(s_n = \dfrac {1}{2^n} + \dfrac {1}{3^n} + \dfrac {1}{4^n} + \dotso\)
\begin{equation*}
\begin{split}
& s_2 = {\color {blue} \dfrac {1}{2^2}} + {\color {red} \dfrac {1}{3^2}} + {\color {brown} \dfrac {1}{4^2}} + \dotso \\\\
& s_3 = {\color {blue} \dfrac {1}{2^3}} + {\color {red} \dfrac {1}{3^3}} + {\color {brown}\dfrac {1}{4^3}} + \dotso \\\\
& s_4 = {\color {blue} \dfrac {1}{2^4}} + {\color {red}\dfrac {1}{3^4}} + {\color {brown}\dfrac {1}{4^4}} + \dotso \quad (+) \\\\
\hline \\
& s_2 + s_3 + s_4 + \dotso = {\color {blue}\dfrac {1}{2^2} + \dfrac {1}{2^3} + \dfrac {1}{2^4} + \dotso} + {\color {red}\dfrac {1}{3^2} + \dfrac {1}{3^3} + \dfrac {1}{3^4} + \dotso} + {\color {brown}\dfrac {1}{4^2} + \dfrac {1}{4^3} + \dfrac {1}{4^4} + \dotso}
\end{split}
\end{equation*}
Deret geometri \({\color {blue}\dfrac {1}{2^2} + \dfrac {1}{2^3} + \dfrac {1}{2^4} + \dotso}\)
\begin{equation*}
\begin{split}
& S_{\sim} = \frac {a}{1 - r} \\\\
& S_{\sim} = \frac {\dfrac {1}{2^2}}{1 - \dfrac 12} \\\\
& S_{\sim} = \frac 12
\end{split}
\end{equation*}
Deret geometri \({\color {red}\dfrac {1}{3^2} + \dfrac {1}{3^3} + \dfrac {1}{3^4} + \dotso}\)
\begin{equation*}
\begin{split}
& S_{\sim} = \frac {a}{1 - r} \\\\
& S_{\sim} = \frac {\dfrac {1}{3^2}}{1 - \dfrac 13} \\\\
& S_{\sim} = \dfrac {1}{3^2} \:.\: \frac 32 \\\\
& S_{\sim} = \frac {1}{2 \:.\: 3}
\end{split}
\end{equation*}
Deret geometri \({\color {brown}\dfrac {1}{4^2} + \dfrac {1}{4^3} + \dfrac {1}{4^4} + \dotso}\)
\begin{equation*}
\begin{split}
& S_{\sim} = \frac {a}{1 - r} \\\\
& S_{\sim} = \frac {\dfrac {1}{4^2}}{1 - \dfrac 14} \\\\
& S_{\sim} = \dfrac {1}{4^2} \:.\: \frac 43 \\\\
& S_{\sim} = \frac {1}{3 \:.\: 4}
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& S_{\text{total}} = \frac {1}{1 \:.\: 2} + \frac {1}{2 \:.\: 3} + \frac {1}{3 \:.\: 4} + \dotso \\\\
& S_{\text{total}} = \sum_1^{\sim} \frac {1}{n \:.\: (n + 1)} \\\\
& S_{\text{total}} = \sum_1^{\sim} \left(\frac 1n - \frac {1}{n + 1} \right)
\end{split}
\end{equation*}
Metode sum of difference
\begin{equation*}
\begin{split}
& S_{1} = \frac 11 \cancel {- \frac 12} \\\\
& S_{2} = \cancel {\frac 12} \cancel {- \frac 13} \\\\
& S_{3} = \cancel {\frac 13} \cancel {- \frac 14} \\\\
& \dotso \\\\
& S_{n} = \cancel {\frac 1n} - \frac {1}{n + 1} \quad (+) \\\\
& \hline \\
& S_{\text{total}} = 1 - \frac {1}{n + 1} \\\\
& S_{\text{total}} = 1 - \frac {1}{\sim} \\\\
& S_{\text{total}} = 1 - 0 \\\\
& S_{\text{total}} = 1
\end{split}
\end{equation*}
Soal 08
SIMAK UI 2014 Matematika Dasar 511
Diketahui deret aritmetika terdiri dari n suku. Suku awal deret tersebut merupakan jumlah n suku pertama bilangan genap dan bedanya n, maka jumlah deret aritmetika tersebut adalah ...
(A) \(n^3\)
(B) \(\dfrac {n^2}{2}\)
(C) \(\dfrac {3n^3}{2} + \dfrac {n^2}{2}\)
(D) \(\dfrac {3n^3}{2} - \dfrac {n^2}{2}\)
(E) \(n^2\)
Jawab: C
Suku pertama
\begin{equation*}
\begin{split}
& a = 2 + 4 + 6 + \dotso + 2n \\\\
& a = \frac n2 [2 + 2n] \\\\
& a = n + n^2
\end{split}
\end{equation*}
Jumlah deret
\begin{equation*}
\begin{split}
& S_n = \frac n2 [2a + (n - 1) \:.\: b] \\\\
& S_n = \frac n2 [2(n + n^2) + (n - 1) \:.\: n] \\\\
& S_n = \frac n2 [2n + 2n^2 + n^2 - n] \\\\
& S_n = \frac n2 [3n^2 + n] \\\\
& S_n = \frac 32 n^3 + \frac 12 n^2
\end{split}
\end{equation*}
Soal 09
SIMAK UI 2015 Matematika Dasar 541
Misalkan tiga suku pertama dari barisan aritmetika adalah \(\log a^3 \: b^7, \: \log a^5 \: b^{12}, \: \log a^8 \: b^{15}\) dan suku ke-12 adalah \(\log a^m \: b^n\). Nilai dari \(2m + n\) adalah ...
(A) 40
(B) 56
(C) 76
(D) 112
(E) 143
Jawab: D
Barisan aritmetika \(\log a^3 \: b^7, \: \log a^5 \: b^{12}, \: \log a^8 \: b^{15}\)
\begin{equation*}
\begin{split}
b & = U_2 - U_1 \\\\
b & = \log a^5 \: b^{12} - \log a^3 \: b^7 \\\\
b & = \log \left(\frac {a^5 \: b^{12}}{a^3 \: b^7} \right) \\\\
b & = \log a^2 \: b^5
\end{split}
\end{equation*}
Suku ke-12
\begin{equation*}
\begin{split}
U_n & = a + (n - 1) \:.\: b \\\\
U_{12} & = \log a^3 \: b^7 + (12 - 1) \:.\: \log a^2 \: b^5 \\\\
U_{12} & = \log a^3 \: b^7 + 11 \:.\: \log a^2 \: b^5 \\\\
U_{12} & = \log a^3 \: b^7 + \log \left(a^2 \: b^5\right)^{11} \\\\
U_{12} & = \log a^3 \: b^7 + \log a^{22} \: b^{55} \\\\
U_{12} & = \log \left( a^3 \: b^7 \:.\: a^{22} \: b^{55} \right) \\\\
U_{12} & = \log a^{25} \: b^{62}
\end{split}
\end{equation*}
\(m = 25, n = 62\)
\(2m + n = 112\)
Soal 10
SIMAK UI 2015 Matematika Dasar 541
Diketahui \(u_n\) dan \(v_n\) adalah barisan aritmetika dengan \(n > 0\). Jumlah n suku pertama dari masing-masing barisan ini adalah \(S_u (n)\) dan \(S_v (n)\). Jika \(\dfrac {S_v (n)}{S_u (n)} = \dfrac {2n + 8}{5n + 9}\) dan \(v_2 = \dfrac 73\), maka \(u_4 = \dotso\)
(A) \(\dfrac {22}{3}\)
(B) \(\dfrac {17}{3}\)
(C) \(4\)
(D) \(\dfrac {11}{3}\)
(E) \(3\)
Soal 11
SIMAK UI 2016 Matematika Dasar 541
Nilai dari \(1 + 2 \:.\: 2 + 3 \:.\: 2^2 + 4 \;.\: 2^3 + \dotso + 2016 \:.\: 2^{2015}\) adalah ...
(A) \(2016 \:.\: 2^{2015}\)
(B) \(2016 \:.\: 2^{2015} + 1\)
(C) \(2015 \:.\: 2^{2016}\)
(D) \(2015 \:.\: 2^{2016} + 1\)
(E) \(2015 \:.\: 2^{2015} - 1\)
Jawab: D
\begin{equation*}
\begin{split}
m & = 1 + {\color {blue} 2 \:.\: 2} + {\color {red}3 \:.\: 2^2} + {\color {brown}4 \;.\: 2^3} + \dotso + 2015 \:.\: 2^{2014} + {\color {green}2016 \:.\: 2^{2015}} \\\\
2m & = {\color {blue}1 \:.\: 2} + {\color {red}2 \:.\: 2^2} + {\color {brown}3 \:.\: 2^3} + 4 \;.\: 2^4 + \dotso + {\color {green}2015 \:.\: 2^{2015}} + 2016 \:.\: 2^{2016} \quad (-) \\\\
\hline \\
-m & = 1 + {\color {blue}2} + {\color {red}2^2} + {\color {brown}2^3} + \dotso + {\color {green}2^{2015}} - 2016 \:.\: 2^{2016} \quad {\color {red} \dotso \: (1)}
\end{split}
\end{equation*}
Bentuk \({\color {blue}2} + {\color {red}2^2} + {\color {brown}2^3} + \dotso + {\color {green}2^{2015}}\) merupakan deret geometri
\(a = 2, \: r = 2, \: n = 2015\)
\begin{equation*}
\begin{split}
S_n & = \frac {a \left(r^n - 1 \right)}{r - 1} \\\\
S_{2015} & = \frac {2 \left(2^{2015} - 1 \right)}{2 - 1} \\\\
S_{2015} & = 2 \left(2^{2015} - 1 \right) \\\\
S_{2015} & = 2^{2016} - 2
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& -m = 1 + {\color {blue}2} + {\color {red}2^2} + {\color {brown}2^3} + \dotso + {\color {green}2^{2015}} - 2016 \:.\: 2^{2016} \quad {\color {red} \dotso \: (1)} \\\\
& -m = 1 + 2^{2016} - 2 - 2016 \:.\: 2^{2016} \\\\
& m = 2016 \:.\: 2^{2016} - 2^{2016} + 1 \\\\
& \bbox[5px, border: 2px solid magenta] {m = 2015 \:.\: 2^{2016} + 1}
\end{split}
\end{equation*}
Soal 12
SIMAK UI 2017 Matematika Dasar 541
Jika diberikan barisan \(4, 8, 14, 22, 32, \dotso\), maka suku ke-20 dari barisan tersebut adalah ...
(A) 382
(B) 392
(C) 402
(D) 412
(E) 422
Jawab: E
Barisan \(4, 8, 14, 22, 32, \dotso\) merupakan barisan aritmetika tingkat 2
\begin{equation*}
\begin{split}
& U_n = a + (n - 1) \:.\: b + (n - 1)(n - 2) \:.\: \dfrac {c}{2!} \\\\
& U_{20} = 4 + (20 - 1) \:.\: 4 + (20 - 1)(20 - 2) \:.\: \dfrac {2}{2!} \\\\
& U_{20} = 4 + 19 \:.\: 4 + 19 \:.\: 18 \:.\: 1 \\\\
& \bbox[5px, border: 2px solid magenta] {U_{20} = 422}
\end{split}
\end{equation*}
Soal 13
SIMAK UI 2018 Matematika Dasar 631
Sembilan buah bilangan membentuk deret aritmetika dan mempunyai jumlah 153. Jika pada setiap dua suku yang berturutan pada deret tersebut disisipkan rata-rata dari 2 suku tersebut, jumlah deret yang baru adalah ...
(A) 267
(B) 279
(C) 289
(D) 315
(E) 349
Jawab: C
Misalkan deret aritmetika tersebut adalah
\(a - 4b, \: a - 3b, \: a - 2b, \: a - b, \: a, \: a + b, \: a + 2b, \: a + 2b, \: a + 3b, \: a + 4b\)
Jumlah bilangan adalah 153
\begin{equation*}
\begin{split}
& a - 4b + a - 3b + a - 2b + a - b + a + a + b + a + 2b + a + 2b + a + 3b + a + 4b = 153 \\\\
& 9a = 153 \\\\
& a = 17
\end{split}
\end{equation*}
Sisipan
\(a - 4b, \: {\color {blue} a - 3\frac 12 b}, \: a - 3b, {\color {blue} a - 2\frac 12 b}, \: a - 2b, \dotso\)
Banyaknya sisipan adalah 8 bilangan
\(a - 3\frac 12 b, \: a - 2\frac 12 b, \: a - 1\frac 12 b, \: a - \frac 12 b, \: a + \frac 12 b, \: a + 1\frac 12 b, \: a + 2\frac 12 b, \: a + 3\frac 12 b\)
Jumlah seluruh sisipan
\begin{equation*}
\begin{split}
& a - 3\tfrac 12 b + a - 2\tfrac 12 b + a - 1\tfrac 12 b + a - \tfrac 12 b + a + \tfrac 12 b + a + 1\tfrac 12 b + a + 2\tfrac 12 b + a + 3\tfrac 12 b \\\\
& 8a \\\\
& 136
\end{split}
\end{equation*}
Jumlah deret baru
\(153 + 136 = \bbox[5px, border: 2px solid magenta] {289}\)
Soal 14
SIMAK UI 2019 Matematika Dasar 521
Jika \(a^2 - bc, \: b^2 - ac, \: c^2 - ab\) adalah barisan aritmetika dengan \(a + b + c = 18\), nilai \(\dfrac {a + c}{b}\) adalah ...
(A) 2
(B) 3
(C) 4
(D) 6
(E) 9
Jawab: A
Barisan aritmetika \(a^2 - bc, \: b^2 - ac, \: c^2 - ab\)
\begin{equation*}
\begin{split}
U_2 - U_1 & = U_3 - U_2 \\\\
b^2 - ac - \left(a^2 - bc \right) & = c^2 - ab - \left(b^2 - ac\right) \\\\
b^2 - ac - a^2 + bc & =c^2 - ab - b^2 + ac \\\\
b^2 - a^2 + bc - ac & = c^2 - b^2 + ac - ab \\\\
(b - a)(b + a) + c(b - a) & = (c - b)(c + b) + a(c - b) \\\\
(b - a) (b + a + c) & = (c - b)(c + b + a) \\\\
(b - a) \cancel{(b + a + c)} & = (c - b) \cancel{(c + b + a)} \\\\
b - a & = c - b \\\\
2b & = a + c
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
& \frac {a + c}{b} = \frac {2b}{b} \\\\
& \bbox[5px, border: 2px solid magenta] {\frac {a + c}{b} = 2}
\end{split}
\end{equation*}
