Recurrence Equations for Arithmetic, Geometric and Differences Progressions

Basic 3 Patterns

$$a_{n+1}-a_n=d$$

This is the recurrence equation of Arithmetic Progression

$$a_1=1,\ \ a_{n+1}-a_n=3 $$

If we write the sequence

$$1,\ 4,\ 7,\ 10,\ 13,\ 16,\ …$$

Formula for $a_n$ of an arithmetic sequence is

$$a_n=a_1+(n-1)d$$

So $a_n$ for this Example Question is

$$a_n= 1+(n-1)(3)$$

$$a_n=3n-2$$

In the question for the recurrence equation, only you get is

$$a_{n+1}-a_n=3$$

you need to imagine the sequence${1,\ 4,\ 7,\ 10,\ 13,\ 16,\ …}$, then you have to use the formula of Arithmetic Progression to get the $a_n$

$$a_n=3n-2$$

The recurrence equation of a Geometric Progression is

$$a_{n+1}=a_n r$$

$$a_1=1,\ \ a_{n+1}=2 a_n$$

We need to imagine that this sequence is like

$$1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ …$$

Then use the formula of Geometric Equation to get the $a_n$ of it

$$a_n=a_1 r^{n-1}$$

$$a_n=2^{n-1}$$

$$a_{n+1}-a_n=b_n$$

$$a_1=6,\ a_{n+1}-a_n=2n+3$$

If you see $a_{n+1}-a_n=b_n$, it is differences progression

Since

$$b_n=2n+3$$

$${b_n}=5,\ 7,\ 9,\ 11,\ 13,\ …$$

Use the formula for $a_n$ of a differences progression

$$a_n=a_1+\sum^{n-1}_{k=1} b_k \ (n≧2)$$

when $n=1, a_n=a_1$

$$a_n=6 +
\sum^{n-1}_{k=1} (2k+3) $$

$$=6+2\bigg( \frac{(n-1)n}{2}\bigg)+ 3(n-1)$$

Remember

$$\sum^{n}_{k=1} k=\frac{n(n+1)}{2}$$

$$\sum^{n}_{k=1} 3=3n $$

$$=6+(n-1)n+3(n-1)$$

$$=n^2+2n+3$$

Since when $ n=1,\ a_1= 6$

$$a_n=n^2+2n+3 $$