3,5 1/2, 10 1/2, 20 1/2,...
find the difference
write a recursive rule and a explict rule
3,5 1/2, 10 1/2, 20 1/2,...
find the difference
write a recursive rule and a explict rule
\(\begin{array}{rrrr} 3,& 5 \dfrac12,& 10 \dfrac12,& 20 \dfrac12,~\cdots\\ \end{array}\)
find the difference:
\(\small{ \begin{array}{rrrrr} n & a(n) & 1.~ \text{difference} & 2.~ \text{difference} & 3.~ \text{difference} \\ \\ \hline \\ 0 & a_0 = 3= \frac62 &\\ & &d_1= \frac52 \\ 1 & a_1 = \frac{11}2 & &d_2= \frac52 \\ & & \frac{10}2 & &d_3= \frac52 \\\\ 2 & a_2 = \frac{21}2 & & \frac{10}2 \\ & & \frac{20}2 \\ 3 & a_3 = \frac{41}2 & &\\ ~\cdots\\ \end{array} }\)
explict rule:
\(\small{ \begin{array}{rcl} a_n &=& \binom{n}{0}a_0 + \binom{n}{1}\cdot d_1 + \binom{n}{2}\cdot d_2 + \binom{n}{3}\cdot d_3\\\\ a_n &=& \binom{n}{0}a_0 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52 \qquad | \qquad \binom{n}{0} = 1 \\\\ a_n &=& a_0 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52 \qquad | \qquad a_0 = 3 \\\\ a_n &=& 3 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52\\ a_n &=& 3 + \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \qquad | \qquad \boxed{ \binom{n}{m} \qquad \text{ if } n<m \rightarrow \binom{n}{m} = 0} \\ \end{array} }\)
recursive rule:
\(\small{ \begin{array}{rcl} a_n &=& 3 + \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1} &=& 3 + \frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right]\\ \hline a_{n+1}-a_n &=&3 + \frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right] - 3 - \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right] - \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n+1}{1}-\binom{n}{1} + \binom{n+1}{2}-\binom{n}{2} + \binom{n+1}{3}- \binom{n}{3} \right]\\\\ \binom{n+1}{1}-\binom{n}{1} &=& \binom{n}{0} \\ \binom{n+1}{2}-\binom{n}{2} &=& \binom{n}{1} \\ \binom{n+1}{3}-\binom{n}{3} &=& \binom{n}{2} \\\\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} \right]\\\\ a_{n+1} &=&a_n + \frac52 \cdot \left[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} \right] \qquad | \qquad \boxed{ \binom{n}{m} \qquad \text{ if } n<m \rightarrow \binom{n}{m} = 0} \\\\\\ \end{array} }\)
the successive differences are
2 1/2, 5, 10
and you can see that these are doubling each time
\(\mbox{A recursive rule would be} \\ a[n] = a[n-1] + \left(\dfrac 5 2\right)\times 2^{n-1} ; a[0]=3\)
\(\mbox{An explicit rule would be} \\ a[n] = \dfrac {1+5 \times 2^n}{2}\)
3,5 1/2, 10 1/2, 20 1/2,...
find the difference
write a recursive rule and a explict rule
\(\begin{array}{rrrr} 3,& 5 \dfrac12,& 10 \dfrac12,& 20 \dfrac12,~\cdots\\ \end{array}\)
find the difference:
\(\small{ \begin{array}{rrrrr} n & a(n) & 1.~ \text{difference} & 2.~ \text{difference} & 3.~ \text{difference} \\ \\ \hline \\ 0 & a_0 = 3= \frac62 &\\ & &d_1= \frac52 \\ 1 & a_1 = \frac{11}2 & &d_2= \frac52 \\ & & \frac{10}2 & &d_3= \frac52 \\\\ 2 & a_2 = \frac{21}2 & & \frac{10}2 \\ & & \frac{20}2 \\ 3 & a_3 = \frac{41}2 & &\\ ~\cdots\\ \end{array} }\)
explict rule:
\(\small{ \begin{array}{rcl} a_n &=& \binom{n}{0}a_0 + \binom{n}{1}\cdot d_1 + \binom{n}{2}\cdot d_2 + \binom{n}{3}\cdot d_3\\\\ a_n &=& \binom{n}{0}a_0 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52 \qquad | \qquad \binom{n}{0} = 1 \\\\ a_n &=& a_0 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52 \qquad | \qquad a_0 = 3 \\\\ a_n &=& 3 + \binom{n}{1}\cdot \frac52 + \binom{n}{2}\cdot \frac52 + \binom{n}{3}\cdot \frac52\\ a_n &=& 3 + \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \qquad | \qquad \boxed{ \binom{n}{m} \qquad \text{ if } n<m \rightarrow \binom{n}{m} = 0} \\ \end{array} }\)
recursive rule:
\(\small{ \begin{array}{rcl} a_n &=& 3 + \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1} &=& 3 + \frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right]\\ \hline a_{n+1}-a_n &=&3 + \frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right] - 3 - \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n+1}{1} + \binom{n+1}{2} + \binom{n+1}{3} \right] - \frac52 \cdot \left[ \binom{n}{1} + \binom{n}{2} + \binom{n}{3} \right] \\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n+1}{1}-\binom{n}{1} + \binom{n+1}{2}-\binom{n}{2} + \binom{n+1}{3}- \binom{n}{3} \right]\\\\ \binom{n+1}{1}-\binom{n}{1} &=& \binom{n}{0} \\ \binom{n+1}{2}-\binom{n}{2} &=& \binom{n}{1} \\ \binom{n+1}{3}-\binom{n}{3} &=& \binom{n}{2} \\\\ a_{n+1}-a_n &=&\frac52 \cdot \left[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} \right]\\\\ a_{n+1} &=&a_n + \frac52 \cdot \left[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} \right] \qquad | \qquad \boxed{ \binom{n}{m} \qquad \text{ if } n<m \rightarrow \binom{n}{m} = 0} \\\\\\ \end{array} }\)