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The series is a(n) = 4n² +5n. . . and this is the general formula which also gives the nth term.

When n = 1, 4n² + 5n = 4*1 + 5*1 = 9

When n = 2, 4n² + 5n = 4*4 + 5*2 = 26 . . .and so on

When n = 5, 4n² + 5n = 4*25 + 5*5 = 125.

The series can be determined using the Difference Technique.

A second difference of 8 gives the first term 4n². . . . as a 2nd difference of 2 equates to n².

After applying this to the original series, the revised series produces a 1st difference of 5, hence 5n.

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Q: What is the nth term of 9 26 51 84 125?
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