#4#, #29#, #129#, #354#, #754#....?

1 Answer
Jun 17, 2018

#a_n = 25n^2-50n+29#

Explanation:

Assuming you are looking for a formula for the #n#th term of this sequence...

First note that no sequence is determined by its first few terms unless you have other information about the sequence - e.g. that it is arithmetic, geometric, quadratic.

We can attempt to find a matching formula by looking at differences between consecutive terms.

Write down the given sequence:

#4, 29, 129, 354, 754#

Write down the sequence of differences between consecutive terms:

#25, 100, 225, 400#

Hmmm. Those look suspiciously like the squares of consecutive multiples of #5#. Note that:

#25=5^2#, #100 = 10^2#, #225=15^2#, #400=20^2#

Hence we can find a matching formula:

#a_n = 4 + (5(n-1))^2#

#color(white)(a_n) = 25n^2-50n+29#