# What is the formula for the #n#th term of the sequence #1, 3, 9, 19,...# ?

##### 1 Answer

#### Answer:

#### Explanation:

Let us examine the differences between successive pairs of terms.

Write down the given sequence:

#color(blue)(1), 3, 9, 19#

Write down the sequence of differences between successive pairs of terms:

#color(blue)(2), 6, 10#

Write down the sequence of difference between successive pairs of those terms:

#color(blue)(4), 4#

Having arrived at a constant sequence after two steps, we can use the first term of each of the above sequences as coefficients to write down a quadratic expression that matches the given sequence:

#a_n = color(blue)(1)/(0!)+color(blue)(2)/(1!)(n-1)+color(blue)(4)/(2!)(n-1)(n-2)#

#color(white)(a_n) = 1+2n-2+2n^2-6n+4#

#color(white)(a_n) = 2n^2-4n+3#

**Footnote**

Here are some more details as to where the formula above comes from.

Given a sequence:

#1, 3, 9, 19#

we can try matching it term by term as follows:

If we only need to match the first term, then we could write:

#a_n = color(blue)(1)#

To match both the first and the second terms, we need to add a multiple of

#a_n = color(blue)(1) + color(blue)(2)(n-1)#

To match the third term too, we need to add a multiple of

#a_n = color(blue)(1) + color(blue)(2)(n-1) + color(blue)(4)/2(n-1)(n-2)#

That formula is sufficient to match the given quadratic sequence.

In general the multiplier for