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Jawab pakai penjelasan​

Berikut ini adalah pertanyaan dari tabraniim43 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Jawab pakai penjelasan​
Jawab pakai penjelasan​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:

3

Penjelasan dengan langkah-langkah:

\large\text{$\begin{aligned}&&S_{255}&=\frac{1}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt[4]{1}+\sqrt[4]{2}\right)}+\frac{1}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[4]{2}+\sqrt[4]{3}\right)}+{}\\&&&{\:\:\vdots\quad}\frac{1}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt[4]{3}+\sqrt[4]{4}\right)}+\dots+\frac{1}{\left(\sqrt{255}+\sqrt{256}\right)\left(\sqrt[4]{255}+\sqrt[4]{256}\right)}\\\\&&&\quad\normalsize\textsf{..... [ tukar posisi sehingga akar yang lebih besar di depan ]}\end{aligned}$}

\large\text{$\begin{aligned}&&S_{255}&=\frac{1}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt[4]{2}+\sqrt[4]{1}\right)}+\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt[4]{3}+\sqrt[4]{2}\right)}+{}\\&&&{\:\:\vdots\quad}\frac{1}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt[4]{4}+\sqrt[4]{3}\right)}+\dots+\frac{1}{\left(\sqrt{256}+\sqrt{255}\right)\left(\sqrt[4]{256}+\sqrt[4]{255}\right)}\end{aligned}$}

\large\text{$\begin{aligned}&&S_{255}&=\sum\limits_{n=1}^{255}\:\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt[4]{n+1}+\sqrt[4]{n}\right)}\end{aligned}$}

Kita perhatikan dan olah rumus suku ke-n-nya.

\large\text{$\begin{aligned}&U_n=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt[4]{n+1}+\sqrt[4]{n}\right)}\\&{\iff}U_n=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt[4]{n+1}+\sqrt[4]{n}\right)}\times\frac{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}\end{aligned}$}

\large\text{$\begin{aligned}&{\qquad}\normalsize\text{$\left[\ \begin{aligned}\left(\sqrt[4]{n+1}+\sqrt[4]{n}\right)\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)&=\left(\sqrt[4]{n+1}\right)^2-\left(\sqrt[4]{n}\right)^2\\&=\left(\sqrt{n+1}-\sqrt{n}\right)\end{aligned}\right.$}\\\\&{\iff}U_n=\frac{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\end{aligned}$}

\large\text{$\begin{aligned}&{\qquad}\normalsize\text{$\left[\ \begin{aligned}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)&=\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\\&=n+1-n\\&=n-n+1\\&=1\end{aligned}\right.$}\\\\&{\iff}U_n=\frac{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}{1}\\\\&{\iff}\boxed{\ \bf U_n=\sqrt[4]{n+1}-\sqrt[4]{n}\ }\end{aligned}$}

Sehingga penjumlahan (deret) di atas menjadi:

(perhatikan indeksnya)

\large\text{$\begin{aligned}&S_{255}=\sum\limits_{n=1}^{255}\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)\\&{\quad\:\;\,}=\sum\limits_{n=1}^{255}\sqrt[4]{n+1}-\sum\limits_{n=1}^{255}\sqrt[4]{n}\\&{\quad\:\;\,}=\sum\limits_{n=\bf2}^{\bf256}\sqrt[4]{n}-\left(\sum\limits_{n=\bf2}^{\bf255}\sqrt[4]{n}+\sum\limits_{n=\bf1}^{\bf1}\sqrt[4]{n}\right)\end{aligned}$}

\large\text{$\begin{aligned}&S_{255}=\left(\sum\limits_{n=\bf2}^{\bf255}\sqrt[4]{n}+\sum\limits_{n=\bf256}^{\bf256}\sqrt[4]{n}\right)-\left(\sum\limits_{n=\bf2}^{\bf255}\sqrt[4]{n}+\sum\limits_{n=\bf1}^{\bf1}\sqrt[4]{n}\right)\\&{\quad\:\;\,}=\sum\limits_{n=\bf2}^{\bf255}\sqrt[4]{n}-\sum\limits_{n=\bf2}^{\bf255}\sqrt[4]{n}+\sum\limits_{n=\bf256}^{\bf256}\sqrt[4]{n}-\sum\limits_{n=\bf1}^{\bf1}\sqrt[4]{n}\\&{\quad\:\;\,}=0+\sqrt[4]{256}-\sqrt[4]{1}\end{aligned}$}

\large\text{$\begin{aligned}&{\quad\:\;\,}=\sqrt[4]{4^4}-\sqrt[4]{1^4}\\&{\quad\:\;\,}=4-1\\&S_{255}=\bf3\end{aligned}$}

∴  Dengan demikian, nilai dari

\large\text{$\begin{aligned}&\frac{1}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt[4]{1}+\sqrt[4]{2}\right)}+\frac{1}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[4]{2}+\sqrt[4]{3}\right)}+{}\\&{\:\:\vdots\quad}\frac{1}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt[4]{3}+\sqrt[4]{4}\right)}+\dots+\frac{1}{\left(\sqrt{255}+\sqrt{256}\right)\left(\sqrt[4]{255}+\sqrt[4]{256}\right)}\\\\\end{aligned}$}

adalah 3.

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Last Update: Sat, 07 May 22
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