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1. Buktikan bahwa

\[16 < \sum_{k=1}^{80} \frac{1}{\sqrt{k}} < 17\]

2. Diberikan $a,b,c$ dan $d$ merupakan solusi dari

\[x^{2018}-11x+10=0\]

Tentukan 

\[\sum_{n=1}^{2017} ((a^n-b^n)+(c^n+d^n))\]

3. Tentukan nilai dari

\[\frac{1}{2} + \frac{1}{2+4} +\frac{1}{2+4+6}+...+\frac{1}{2+4+6+...+4034}\]

4. Tentukan nilai dari

\[\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{2017}{2015!+2016!+2017!}\]

5. Tentukan nilai dari

\[\sum_{k=1}^{2017} \frac{1}{(k+1)\sqrt{k} + k\sqrt{k+1}}\]

6. Tentukan nilai dari

\[\sum_{k=1}^{2017} \frac{1}{\sqrt{k} + \frac{1}{\sqrt{k} + \frac{1}{...}}}\]

7. Buktikan jika $a > b > 0$ dan

\[x=\frac{1+a+a^2+a^3+...+a^{n-1}}{1+a+a^2+a^3+...+a^n}\]

\[y=\frac{1+b+b^2+b^3+...+b^{n-1}}{1+b+b^2+b^3+...+b^n}\]

maka $x < y$.

8. Diberikan $a_n = \frac{n}{2017}$ untuk $n \in \mathbb{N}$. Tentukan nilai dari

\[\sum_{k=1}^{2017} \frac{a_k^5}{1 + 5a_k^4 - 10a_k^3 + 10a_k^2 - 5a_k}\]

9. Diberikan fungsi $f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}_0$ dimana $\mathbb{N}_0=\mathbb{N} \cup \{0\}$. Jika $f(m+1,n) = f(m,n) + m$ dan $f(m,n+1)=f(m,n) + n$ dimana $m,n \in \mathbb{N}$ serta $f(1,1)=0$, tentukan semua pasangan $(p,q)$ yang memenuhi $f(p,q)=2017$.

10.Diberikan $\triangle ABC$, ditarik garis lurus beruturut-turut dari titik $A,B,C$ dan memotong sisi dihadapannya di titik $F,D,E$ serta ketiga garis tersebut berpotongan di titik $G$. 

Jika panjang $DG=GF=GE$ dan $AG+BG+CG=43$, tentukan $AG \cdot BG \cdot CG$.

IMG_20171105_101204.jpg

Edited by Wildan Bagus W

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No 3

Spoiler

$=\frac{1}{2} + \frac{1}{2 + 4} + \frac{2 + 4 + 6} + ...+ \frac{1}{2+4+6+...+4034$

$\large=\sum_{n=1}^{2017} \frac{1}{2(1+2+3+...+k)}$

$\large=\sum_{n=1}^{2017} \frac{1}{2 \cdot \frac{n(n+1)}{2}}$

$\large=\sum_{n=1}^{2017} \frac{1}{n(n+1)}$

$\large=\sum_{n=1}^{2017} (\frac{1}{n} - \frac{1}{n+1})$

$= 1 - \frac{1}{2017}$

$=\frac{2016}{2017}$

$\large \therefore \sum_{n=1}^{2017} \frac{1}{2+4+6+...+2n} = \frac{2016}{2017}$

 

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No 4

Spoiler

$=\sum_{k=1}^{2015} \frac{k+2}{k! + (k+1)! + (k+2)!}$

$= \sum_{k=1}^{2015} \frac{1}{k!}[\frac{k+2}{1+k+1+(k+1)(k+2)}]$

$=\sum_{k=1}^{2015} \frac{1}{k!}[\frac{k+2}{k+2+(k+1)(k+2)}]$

$=\sum_{k=1}^{2015}\frac{1}{k!}[ \frac{k+2}{(k+2)^2}]$

$= \sum_{k=1}^{2015} \frac{1}{k!} \cdot \frac{1}{k+2)}$

$=\sum_{k=1}^{2015} \frac{k+1}{(k+2)!}$

$=\sum_{k=1}^{2015} \frac{k+2-1}{(k+2)!}$

$= \sum_{k=1}^{2015} (\frac{1}{(k+1)!} - \frac{1}{(k+2)!}$

$=\frac{1}{2!} - \frac{1}{2017!}$

 

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5.

Spoiler

\begin{align*}\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}&=\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}\cdot\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)\sqrt{k}-k\sqrt{k+1}}\\&=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)^2k-k^2(k+1)}\\&=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)}\\&=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}\\&=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\\\sum_{k=1}^{2017}\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}&=\sum_{k=1}^{2017}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\\&=\boxed{1-\frac{1}{\sqrt{2018}}}\end{align*}

 

6.

Spoiler

Misalkan \begin{align*}x&=\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{...}}}\\&=\frac{1}{\sqrt{k}+x}\\x^2+\sqrt{k}x-1&=0\\x&=\frac{-\sqrt{k}+\sqrt{k-4(1)(-1)}}{2(1)}\\&=\frac{\sqrt{k+4}-\sqrt{k}}{2}\end{align*} Maka \begin{align*}\sum_{k=1}^{2017}\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}+\frac{1}{...}}}&=\sum_{k=1}^{2017}\frac{\sqrt{k+4}-\sqrt{k}}{2}\\&=\frac{1}{2}\left\{(\sqrt{5}-\sqrt{1})+(\sqrt{6}-\sqrt{2})+(\sqrt{7}-\sqrt{3})+\dots+(\sqrt{2021}-\sqrt{2017})\right\}\\&=\boxed{\frac{1}{2}\left(-\sqrt{1}-\sqrt{2}-\sqrt{3}-\sqrt{4}+\sqrt{2018}+\sqrt{2019}+\sqrt{2020}+\sqrt{2021}\right)}.\end{align*}

 

7. 

Spoiler

 

\begin{align*}x&=\frac{1+a+a^2+a^3+\dots+a^{n-1}}{1+a+a^2+a^3+\dots+a^n}\\&=\frac{1+a+a^2+a^3+\dots+a^{n-1}}{1+a+a^2+a^3+\dots+a^n}\cdot\frac{1-a}{1-a}\\&=\frac{1-a^n}{1-a^{n+1}}\\&=\frac{a^n-1}{a^{n+1}-1}\\\lim_{n\to\infty}x&=\lim_{n\to\infty}\frac{1-\frac{1}{a^n}}{a-\frac{1}{a^n}}\\&=\frac{1}{a}\end{align*}

\begin{align*}y&=\frac{1+b+b^2+b^3+\dots+b^{n-1}}{1+b+b^2+b^3+\dots+b^n}\\&=\frac{1+b+b^2+b^3+...+b^{n-1}}{1+b+b^2+b^3+...+b^n}\cdot\frac{1-b}{1-b}\\&=\frac{1-b^n}{1-b^{n+1}}\\&=\frac{b^{n}-1}{b^{n+1}-1}\\\lim_{n\to\infty}y&=\lim_{n\to\infty}\frac{b^{n}-1}{b^{n+1}-1}\\&=\lim_{n\to\infty}\frac{1-\frac{1}{b^n}}{b-\frac{1}{b^{n}}}\\&=\frac{1}{b}\end{align*}

Karena $a>b$, maka $x<y$.

 

 

Edited by Fachni

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