+279 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?
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
+129852 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
+26393 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
+26393 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:
\(\huge{ \begin{equation} a_n = \begin{cases} 2000+n \text{,} & \text{if } \ n \ \text{ is odd} \\ 2004-n \text{,} & \text{if } \ n \ \text{ is even} \\ \end{cases} \end{equation} }\)
\(\large{ \begin{array}{|rcll|} \hline a_{2004} &=& 2004 - 2004 \\ &=& 0 \\ \hline \end{array} }\)
