What is the #n#th term of 1, 7, 17, 31...?

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Answer 1 · 2016-10-05T20:32:56

#2n^2-1#

You need to find a pattern between the value and the place in the sequence that value holds: we have

#1 -> 1#
#2 -> 7#
#3 -> 17#
#4 -> 31#

There's no standard way to find such a pattern, you can only have some smart idea by looking at the numbers.

For example, in this case, you may notice that all the numbers are one less than twice a perfect square:

#1 -> 1 = 2*(1^2)-1#
#2 -> 7 = 2*(2^2)-1#
#3 -> 17 = 2*(3^2)-1#
#4 -> 31 = 2*(4^2)-1#

It wasn't immediate, but once you have this, the pattern should be clear: we can predict

#1 -> 2*(1^2)-1#
#2 -> 2*(2^2)-1#
#3 -> 2*(3^2)-1#
#4 -> 2*(4^2)-1#
#5 -> 2*(5^2)-1#
#5 -> 2*(5^2)-1#

and thus, in general, #n -> 2n^2-1#