+0  
 
0
112
4
avatar+211 

In the sequence {2001, 2002, 2003,...} each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001+2002-2003=2000. What is the value of the 2004th term? 

yasbib555  Aug 8, 2018
 #1
avatar
+1

2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, 2009, 1994, 2011, 1992, 2013, 1990, 2015, 1988, 2017............etc.
a_n = (-1)^n (-n + 2002 (-1)^n + 2) 
a(2004) = (-1)^2004(-2004 + 2002(-1)^2004 + 2)
a(2004) = 1(-2004 + 2002(1) + 2)
a(2004) = (-2004 + 2004)
a(2004) = 0

Guest Aug 8, 2018
edited by Guest  Aug 8, 2018
 #2
avatar+91160 
+2

Discovering a pattern  is often the key to  solving this type  of problem......

Let's look  at the  first 12  terms....we have

 

( 2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, 2009,  1994,  2011, 1992 .....  )

 

Note that the patern appears to  be that every  4th term decreases the series  by 4 

 

So we have that

 

2000  - 0(4)  =  2000  - 0   = 4th term

2000 - 1(4)  = 2000- 4 =  1996  =  8th term

2000 - 2(4)  = 2000 - 8 =  1992  = 12th term

 

 

So

2000 - 500(4) =2000 - 2000  = 0 =   2004th term 

 

 

cool cool cool

CPhill  Aug 8, 2018
 #3
avatar+20153 
0

In the sequence {2001, 2002, 2003,...} each term after the third is found

by subtracting the previous term from the sum of the two terms that precede that term.

For example, the fourth term is 2001+2002-2003=2000.

What is the value of the 2004th term? 

 

\(\begin{array}{|lcll|} \hline a_1 &=& 2001 \\ a_2 &=& 2002 \\ a_3 &=& 2003 \\ a_4 &=& -1\cdot a_3 + 1\cdot a_2 + 1\cdot a_1 \\ a_5 = -1\cdot a_4 + 1\cdot a_3 + 1\cdot a_2 &=& +2\cdot a_3 + 0\cdot a_2 - 1\cdot a_1 \\ a_6 = -1\cdot a_5 + 1\cdot a_4 + 1\cdot a_3 &=& -2\cdot a_3 + 1\cdot a_2 + 2\cdot a_1 \\ a_7 = -1\cdot a_6 + 1\cdot a_5 + 1\cdot a_4 &=& +3\cdot a_3 + 0\cdot a_2 - 2\cdot a_1 \\ a_8 = -1\cdot a_7 + 1\cdot a_6 + 1\cdot a_5 &=& -3\cdot a_3 + 1\cdot a_2 + 3\cdot a_1 \\ a_9 = -1\cdot a_8 + 1\cdot a_7 + 1\cdot a_6 &=& +4\cdot a_3 + 0\cdot a_2 - 3\cdot a_1 \\ a_{10} = -1\cdot a_9 + 1\cdot a_8 + 1\cdot a_7 &=& -4\cdot a_3 + 1\cdot a_2 + 4\cdot a_1 \\ a_{11} = -1\cdot a_{10} + 1\cdot a_{9} + 1\cdot a_{8} &=& +5\cdot a_3 + 0\cdot a_2 - 4\cdot a_1 \\ a_{12} = -1\cdot a_{11} + 1\cdot a_{10} + 1\cdot a_{9} &=& -5\cdot a_3 + 1\cdot a_2 + 5\cdot a_1 \\ a_{13} = -1\cdot a_{12} + 1\cdot a_{11} + 1\cdot a_{10} &=& +6\cdot a_3 + 0\cdot a_2 - 5\cdot a_1 \\ a_{14} = -1\cdot a_{13} + 1\cdot a_{12} + 1\cdot a_{11} &=& -6\cdot a_3 + 1\cdot a_2 + 6\cdot a_1 \\ \ldots \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_n} & \mathbf{=} & \mathbf{ -(-1)^n\lfloor \dfrac{n-1}{2}\rfloor \cdot a_3 + \lfloor\dfrac{1+(-1)^n}{2}\rfloor \cdot a_2+(-1)^n\lfloor \dfrac{n-2}{2} \rfloor \cdot a_1 } \\\\ a_{2014} &=& \small{ -(-1)^{2014}\lfloor \dfrac{2014-1}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+(-1)^{2014}}{2}\rfloor \cdot 2002 +(-1)^{2014}\lfloor \dfrac{2014-2}{2} \rfloor \cdot 2001 } \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+1}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+1}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{2}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -1006 \cdot 2003 + 1 \cdot 2002 +1\cdot 1006 \cdot 2001 \\\\ &=& -2015018 + 2002 + 2013006 \\\\ &=& -2013016 + 2013006 \\\\ &=& 0 \\ \hline \end{array} \)

 

The value of the 2004th term is 0

 

laugh

heureka  Aug 9, 2018
edited by heureka  Aug 9, 2018
 #4
avatar+20153 
0

In the sequence {2001, 2002, 2003,...} each term after the third is found

by subtracting the previous term from the sum of the two terms that precede that term.

For example, the fourth term is 2001+2002-2003=2000.

What is the value of the 2004th term? 

 

 

2. Solution:

 

\(\huge{ \begin{equation} a_n = \begin{cases} 2000+n \text{,} & \text{if } \ n \ \text{ is odd} \\ 2004-n \text{,} & \text{if } \ n \ \text{ is even} \\ \end{cases} \end{equation} }\)

 

\(\large{ \begin{array}{|rcll|} \hline a_{2004} &=& 2004 - 2004 \\ &=& 0 \\ \hline \end{array} }\)

 

laugh

heureka  Aug 9, 2018

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