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1) Find the sum of the infinite series \(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots.\)

2) In a certain sequence the first term is \(a_1\) = 2007 and the second term is \(a_2\) = 2008. Furthermore, the values of the remaining terms are chosen so that \(a_n + a_{n + 1} + a_{n + 2} = n\) for all \(n \ge 1\). Determine \(a_{1000}.\)

 Jul 9, 2019
 #1
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+1

1) I don't know how to sum them up mathematically, but my computer comes up with this summation:
sumfor(n, 1, 1000,  ((n+1) * (1 / 1998)^n)) + 1=3992004/3988009 = 1.0010017530 0507....................

 Jul 10, 2019
 #2
avatar+23171 
+2

1)
Find the sum of the infinite series
\(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots\).

 

\(\text{Let $r=\dfrac{1}{1998} $}\)

 

\(\begin{array}{|rcll|} \hline s&=& \mathbf{1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots} \\ s&=& \mathbf{1+2r +3r^2+4r^3+\cdots + nr^{n-1} + \ldots} \\ \hline \end{array} \)

 

\(\begin{array}{|rcrl|} \hline s_n &=& 1+&2r +3r^2+4r^3+\cdots + nr^{n-1} \quad |\quad \cdot r \\ rs_n &=& & 1r +2r^2+3r^3+\cdots + (n-1)r^{n-1}+nr^n \\ \hline s_n-rs_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} -nr^n \\ \hline \end{array}\\ \begin{array}{|rclrcrl|} \hline s_n(1-r) &=& \underbrace{1+r+r^2+r^3+\ldots + r^{n-1}}_{=S_n(\text{ geometric progression})} -nr^n \\ s_n(1-r) &=& S_n -nr^n & S_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} \quad | \quad \cdot r\\ & & & rS_n &=& &r+r^2+r^3+\ldots + r^{n-1}+r^n \\ \hline & & & S_n-rS_n &=& 1- &r^n \\ & & & S_n(1-r) &=& 1- &r^n \\ & & & S_n &=& \dfrac{1- r^n}{1-r} \\ s_n(1-r) &=& \dfrac{1- r^n}{1-r} -nr^n \\ s_n &=& \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ s &=& \lim \limits_{n\to \infty} s_n \\ s &=& \lim \limits_{n\to \infty} \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ && \boxed{\lim \limits_{n\to \infty} r^n = \lim \limits_{n\to \infty} \left(\dfrac{1}{1998}\right)^n = 0} \\ s &=& \dfrac{1- 0}{(1-r)^2} - 0 \\ s &=& \dfrac{1}{(1-r)^2} \\ s &=& \dfrac{1}{\left(1-\dfrac{1}{1998}\right)^2} \\ s &=& \left(\dfrac{1998}{1997}\right)^2 \\ \mathbf{s} &=& \mathbf{1.00100175300507095\ldots...} \\ \hline \end{array}\)

 

laugh

 Jul 10, 2019
edited by heureka  Jul 10, 2019
edited by heureka  Jul 10, 2019
 #3
avatar+23171 
+2

2) 

In a certain sequence the first term is \(a_1= 2007\) and the second term is \(a_2 = 2008\).

Furthermore, the values of the remaining terms are chosen so that \( a_n + a_{n + 1} + a_{n + 2} = n\) for all \(n \ge 1\).

Determine \(a_{1000}\).

 

\(\begin{array}{|lrcll|} \hline (1) & a_n + a_{n + 1} + a_{n + 2} &=& n \\ (2) & a_{n+1} + a_{n + 2} + a_{n + 3} &=& n+1 \\ \hline (2)-(1): & a_{n+1} + a_{n + 2} + a_{n + 3}-(a_n + a_{n + 1} + a_{n + 2}) &=& n+1-n \\ & a_{n+1} + a_{n + 2} + a_{n + 3}-a_n - a_{n + 1} - a_{n + 2} &=& 1 \\ & a_{n + 3}-a_n &=& 1 \\ & \mathbf{a_{n + 3}} &=& \mathbf{1+a_n} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline & \mathbf{a_{n + 3}} &=& \mathbf{1+a_n} \\ \hline n= 1: & a_{1 + 3}=\mathbf{a_{4}} &=& \mathbf{1+a_1} \\ \hline n= 4: & a_{4 + 3}=a_{7} &=& 1+a_4 \\ & &=& 1+( 1+a_1) \\ & \mathbf{ a_{7}} &=& \mathbf{2+a_1} \\ \hline n= 7: & a_{7 + 3}=a_{10} &=& 1+a_7 \\ & &=& 1+( 2+a_1) \\ & \mathbf{ a_{10}} &=& \mathbf{3+a_1} \\ \hline n= 10: & a_{10 + 3}=a_{13} &=& 1+a_{10} \\ & &=& 1+( 3+a_1) \\ & \mathbf{ a_{13}} &=& \mathbf{4+a_1} \\ \hline \ldots \\ \hline \end{array}\)

 

generalized:

\(\begin{array}{|rcll|} \hline \mathbf{a_{4}} &=& \mathbf{1+a_1} \\ \mathbf{ a_{7}} &=& \mathbf{2+a_1} \\ \mathbf{ a_{10}} &=& \mathbf{3+a_1} \\ \mathbf{ a_{13}} &=& \mathbf{4+a_1} \\ \ldots \\ \mathbf{ a_{1+3m}} &=& \mathbf{m+a_1} \\ \hline \end{array} \)

 

\(\mathbf{a_{1000} =\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a_{1+3m}} &=& \mathbf{m+a_1} \\ \hline 1+3m &=& 1000 \\ 3m &=& 999 \\ m &=& 333 \\ a_{1000} &=& 333+a_1 \quad | \quad a_1 =2007 \\ a_{1000} &=& 333 + 2007 \\ \mathbf{a_{1000}} &=& \mathbf{2340} \\ \hline \end{array}\)

 

laugh

 Jul 10, 2019
edited by heureka  Jul 10, 2019

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