What is the formula for the sequence 1, 3, 7, 9,... ?

2 Answers
Sep 20, 2017

n+(n-1)

(assuming that the sequence ie 1,3,5,7,9)

Explanation:

"n of " 2=3
"n of " 3=5
"n of " 4=7
"n of " 5=9

Basically you just work out how different a is from n and adjust to suit. In this case, the pattern is the n plus what the previous n was: n-1.

Checking my idea of the pattern:
If n = 1, then a = 1+(1-1)=1+0=1 which is correct.

If n = 7, then a = 5+(5-1) = 5 + 4 = 9 which is also correct.

If n = k, then a = k +(k-1) = 2k-1 which when k is substituted in is the same as above (2*5=10)

Sep 20, 2017

This sequence can be matched by:

a_n = 3n-5/2-(-1)^n/2

or by:

a_n = -2/3n^3+5n^2-25/3n+5

Explanation:

Assuming that the question is correct as given, there are several way to match the sequence:

1, 3, 7, 9

with a formula.

For example, if we add (-1)^n/2 to each element, then we get the sequence:

1/2, 7/2, 13/2, 19/2

which is an arithmetic sequence with common difference 3.

Hence we can write a formula for the given sequence:

a_n = 3n-5/2-(-1)^n/2

Alternatively, we could use the method of differences to match the sequence with a polynomial...

Write down the given sequence:

color(blue)(1), 3, 7, 9

Write down the sequence of differences between each consecutive pair of terms:

color(purple)(2), 4, 2

Write down the sequence of differences of those differences:

color(brown)(2), -2

Write down the sequence of differences of those differences:

color(green)(-4)

Having arrived at a constant sequence (albeit of just one element), we can use the initial term of each of these sequences as coefficients to give us a direct formula:

a_n = color(blue)(1)/(0!) + color(purple)(2)/(1!)(n-1) + color(brown)(2)/(2!)(n-1)(n-2) + color(green)(-4)/(3!)(n-1)(n-2)(n-3)

color(white)(a_n) = 1+2n-2+n^2-3n+2-2/3n^3+4n^2-22/3n+4

color(white)(a_n) = -2/3n^3+5n^2-25/3n+5