itu soal nya ada di gambar pliss bantuin ya ,

Berikut ini adalah pertanyaan dari anniemilanuraeni pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Itu soal nya ada di gambar pliss bantuin ya , yang bener

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Nilai $a + b - c$ adalah78.
 

Pembahasan

Diketahui

$\begin{aligned}&100-\left(\frac{3\times4}{2\times5}+\frac{6\times7}{5\times8}+\frac{9\times10}{8\times11}+{\dots}+\frac{42\times43}{41\times44}\right)\\&=a\,\frac{b}{c}\end{aligned}$

Ditanyakan

Nilai $a + b - c$

PENYELESAIAN

$\begin{aligned}&100-\overbrace{\left(\frac{3\times4}{2\times5}+\frac{6\times7}{5\times8}+\frac{9\times10}{8\times11}+{\dots}+\frac{42\times43}{41\times44}\right)}^{\begin{array}{c}42/3=14\rm\ suku\end{array}}\\{=\ }&100-\sum_{n=1}^{14}\frac{3n(3n+1)}{(3n-1)(3n+2)}\\{=\ }&100-\sum_{n=1}^{14}\frac{[(3n-)1+1][(3n+2)-1]}{(3n-1)(3n+2)}\\{=\ }&100-\sum_{n=1}^{14}\frac{(3n-1)(3n+2)+(3n+2)-(3n-1)-1}{(3n-1)(3n+2)}\\{=\ }&100-\sum_{n=1}^{14}\frac{(3n-1)(3n+2)+2+1-1}{(3n-1)(3n+2)}\end{aligned}$
$\begin{aligned}{=\ }&100-\sum_{n=1}^{14}\frac{(3n-1)(3n+2)+2}{(3n-1)(3n+2)}\\{=\ }&100-\sum_{n=1}^{14}\left(\frac{(3n-1)(3n+2)}{(3n-1)(3n+2)}+\frac{2}{(3n-1)(3n+2)}\right)\\{=\ }&100-\sum_{n=1}^{14}\left(1+\frac{2}{(3n-1)(3n+2)}\right)\\{=\ }&100-\left(\sum_{n=1}^{14}1+\sum_{n=1}^{14}\frac{2}{(3n-1)(3n+2)}\right)\\{=\ }&100-\left(14\cdot1+\sum_{n=1}^{14}\frac{2}{(3n-1)(3n+2)}\right)\\{=\ }&86-\sum_{n=1}^{14}\frac{2}{(3n-1)(3n+2)}\quad...(\bigstar)\end{aligned}$

Untuk penyelesaian notasi sigma, kita lakukan dekomposisibentuk pecahan tersebut menjadipecahan parsial.

$\begin{aligned}\frac{2}{(3n-1)(3n+2)}&=\frac{a_0}{3n-1}+\frac{a_1}{3n+2}\\\frac{2}{(3n-1)(3n+2)}&=\frac{a_0(3n+2)+a_1(3n-1)}{(3n-1)(3n+2)}\end{aligned}$

Pada pembilang, kita memperoleh:

$a_0(3n+2)+a_1(3n-1)=2$

Pada penyebut, akar-akarnya adalah n = 1/3 dan n = –2/3.

$\begin{aligned}&n=\frac{1}{3}:\\&\ \Rightarrow a_0(1+2)+a_1(0)=2\\&\ \Rightarrow 3a_0=2\\&\ \Rightarrow a_0=\frac{2}{3}\\&n=-\frac{2}{3}:\\&\ \Rightarrow a_0(0)+a_1(-2-1)=2\\&\ \Rightarrow -3a_1=2\\&\ \Rightarrow a_1=-\frac{2}{3}\\\end{aligned}$

Sehingga:

$\begin{aligned}\frac{2}{(3n-1)(3n+2)}&=\frac{2}{3(3n-1)}-\frac{2}{3(3n+2)}\\\Rightarrow \frac{2}{(3n-1)(3n+2)}&=\frac{2}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\end{aligned}$

Kita substitusi persamaan terakhir ke dalam $(\bigstar)$ .

$\begin{aligned}&100-\left(\frac{3\times4}{2\times5}+\frac{6\times7}{5\times8}+\frac{9\times10}{8\times11}+{\dots}+\frac{42\times43}{41\times44}\right)\\{=\ }&86-\sum_{n=1}^{14}\frac{2}{(3n-1)(3n+2)}\\{=\ }&86-\sum_{n=1}^{14}\frac{2}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\\{=\ }&86-\frac{2}{3}\sum_{n=1}^{14}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\\{=\ }&86-\frac{2}{3}\left(\sum_{n=1}^{14}\frac{1}{3n-1}-\sum_{n=1}^{14}\frac{1}{3n+2}\right)\quad...(\bigstar\bigstar)\\\end{aligned}$

Kemudian, kita manipulasi indeks notasi sigma menjadi bentuk lain yang ekuivalen.

$\begin{aligned}3n+2&=3n-1+3\\3n+2&=3(n+1)-1\\\Rightarrow\ \sum_{n=1}^{14}\frac{1}{3n+2}&=\sum_{n=1}^{14}\frac{1}{3(n+1)-1}\\\therefore\ \sum_{n=1}^{14}\frac{1}{3n+2}&=\sum_{n=2}^{15}\frac{1}{3n-1}\end{aligned}$

Kita substitusi persamaan terakhir ke dalam $(\bigstar\bigstar)$ .

$\begin{aligned}&100-\left(\frac{3\times4}{2\times5}+\frac{6\times7}{5\times8}+\frac{9\times10}{8\times11}+{\dots}+\frac{42\times43}{41\times44}\right)\\{=\ }&86-\frac{2}{3}\left(\sum_{n=1}^{14}\frac{1}{3n-1}-\sum_{n=1}^{14}\frac{1}{3n+2}\right)\\{=\ }&86-\frac{2}{3}\left(\sum_{n=1}^{14}\frac{1}{3n-1}-\sum_{n=2}^{15}\frac{1}{3n-1}\right)\end{aligned}$
$\begin{aligned}{=\ }&86-\frac{2}{3}\left(\sum_{n=1}^{1}\frac{1}{3n-1}+\sum_{n=2}^{14}\frac{1}{3n-1}-\left(\sum_{n=2}^{14}\frac{1}{3n-1}+\sum_{n=15}^{15}\frac{1}{3n-1}\right)\right)\\{=\ }&86-\frac{2}{3}\left(\sum_{n=1}^{1}\frac{1}{3n-1}-\sum_{n=15}^{15}\frac{1}{3n-1}\right)\\{=\ }&86-\frac{2}{3}\left(\frac{1}{3-1}-\frac{1}{45-1}\right)\\{=\ }&86-\frac{2}{3}\left(\frac{1}{2}-\frac{1}{44}\right)\\{=\ }&86-\frac{2}{3}\left(\frac{21}{44}\right)\\{=\ }&86-\frac{7}{22}\end{aligned}$
$\begin{aligned}{=\ }&\boxed{\ \bf85\,\frac{15}{22}\ }\iff\bf a\,\frac{b}{c}\end{aligned}$