what are the next two numbers? 0,1,4,5,8,?,?
I will venture to guess your series: 0, 1, 4, 5, 8, 9, 12, 13, 16.......and so on. I think your series goes like this: add the last 2 terms and subtract the one before.
+26400 what are the next two numbers? 0,1,4,5,8,?,?
\(\begin{array}{rcll} a_1 &=& 0\\ && & +1 \\ a_2 &=& 1\\ && & +3 \\ a_3 &=& 4 \\ && & +1 \\ a_4 &=& 5 \\ && & +3 \\ a_5 &=& 8 \\ && & +1 \\ a_6 &=& 9 \\ && & +3 \\ a_7 &=& 12 \\ && & +1 \\ a_8 &=& 13 \\ && & +3 \\ a_9 &=& 16 \\ && & +1 \\ a_{10} &=& 17 \\ \cdots \end{array}\)
\(\begin{array}{rcll} a_{n+1}= \left\{ \begin{array}{rl} 2\cdot a_n-2 \qquad & (\text{n odd})\\ 2\cdot a_n-3 \qquad & (\text{n even}) \end{array} \right. \end{array} \)
\(\color{red} \boxed{~ \color{black} \begin{array}{rcll} a_n &=& (2\cdot n - 2) -\frac12 [~ 1+(-1)^n ~] \end{array} ~} \)
+26400 what are the next two numbers? 0,1,4,5,8,?,?
without mistakes:
\(\begin{array}{r|rcll} n &\\ \hline 1& a_1 &=& 0\\ & && & +1 \\ 2& a_2 &=& 1\\ & && & +3 \\ 3& a_3 &=& 4 \\ & && & +1 \\ 4& a_4 &=& 5 \\ & && & +3 \\ 5& a_5 &=& 8 \\ & && & +1 \\ 6& a_6 &=& 9 \\ & && & +3 \\ 7& a_7 &=& 12 \\ & && & +1 \\ 8& a_8 &=& 13 \\ & && & +3 \\ 9& a_9 &=& 16 \\ & && & +1 \\ 10& a_{10} &=& 17 \\ \cdots \end{array} \)
\(\begin{array}{rcll} a_{n}= \left\{ \begin{array}{rl} 2\cdot n-2 \qquad & (\text{n odd})\\ 2\cdot n-3 \qquad & (\text{n even}) \end{array} \right. \end{array}\)
\(\color{red} \boxed{~ \color{black} \begin{array}{rcll} a_n &=& (2\cdot n - 2) -\frac12 [~ 1+(-1)^n ~] \end{array} ~}\)
+130514 Thanks for that answer, heureka.........could you explain how you derived your final formula?
+26400 Thanks for that answer, heureka.........
could you explain how you derived your final formula?
...so far:
\(\begin{array}{rcll} a_{n}= \left\{ \begin{array}{ll} 2\cdot n-2 = (2\cdot n - 2) + 0\qquad & (\text{n odd})\\ 2\cdot n-3 = (2\cdot n - 2) -1 \qquad & (\text{n even}) \end{array} \right. \end{array} \)
...composite:
\(\begin{array}{rcll} a_n = (2\cdot n - 2) -1 \cdot s \end{array}\)
s is a switch. We need a function for a switch.
if n is even, s is on or equal 1,
and if n is odd, s is off or equal 0.
As the next we see we can switch with \((-1)^n\)
This function does switch beween -1 and 1 if n is odd or if n is even.
\(\begin{array}{lcr} (-1)^{\text{odd number}} &=& -1 \\ (-1)^{\text{even number}} &=& 1 \end{array} \)
We must modify this "switch" function to get our function.
But it is easy to do this.
First step we add 1.
\(\begin{array}{lcr} 1 + (-1)^{\text{odd number}} &=& -1+1 = 0\\ 1 + (-1)^{\text{even number}}&=& 1+1 = 2\\ \end{array} \)
Second step, the number 2 must be number 1, so we divide by 2
\(\begin{array}{lcr} \dfrac{1 + (-1)^{\text{odd number}} } {2} &=& \frac{0}{2} = 0\\\\ \dfrac{1 + (-1)^{\text{even number}} } {2}&=& \frac{2}{2} = 1\\ \end{array}\)
So our switch s is ready:
\(\begin{array}{lrcr} & s &=& \dfrac{1 + (-1)^n } {2} \\ & s &=& \frac12 [~ 1+(-1)^n ~]\\ n & \text{ is odd} & s = 0 \\ n & \text{ is even} & s = 1 \end{array}\)
