,
,
,
,
,
That has 6 terms, and we don't see an obvious pattern for it.
1 less than 6 is 5, so we assume a 5th degree polynomial equation
in n for the nth term
.
That is, we assume the 5th degree polynomial
We substitute 1,2,3,4,5, and 6 for n, making 6 equations
in 6 unknowns:
Simplifying:
Write the augmented matrix:
If you are using a TI-83 or TI-84, put that
in matrix [A], then in the 2nd MATRIX MATH
menu select rref([A]} and get
You could do that without a calculator using row
operations, but it would take you a long time and
would be tedious
Change all those decimals to fractions:
So that matrix gives us
So the equation that will produce
,
,
,
,
,
is found by substituting those values in
and if you like you can find as many terms as you like by substituting
n=7,8,9,10,11,12,...:
,
,
,
,
,
, etc.
Edwin