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(9n2 - 121) = (3n + 11) (3n - 11)
3n(n+1] + 5 is the nth term
5
The Series has the formula 3n + 1/2(n - 1)(n - 2) = 3n + 1/2(n2 - 3n + 2) which simplifies to, 1/2(n2 +3n + 2)
3n and n = 4? 3n ( the four replaces the n 3(4) = 12 =======
3n^4 divided by 3n^3 = n
When n=30, 3n-1 = 89 .
Just plug in 30 for n in 3n-1. The answer is 89.
(9n2 - 121) = (3n + 11) (3n - 11)
3n(n+1] + 5 is the nth term
t(n) = 28-3n where n = 1,2,3,...
5
The Series has the formula 3n + 1/2(n - 1)(n - 2) = 3n + 1/2(n2 - 3n + 2) which simplifies to, 1/2(n2 +3n + 2)
t(n) = 3n2 + n = n(3n + 1)
3n - 2n = n 3n - 2n = (3 - 2)n = 1n = n
3n and n = 4? 3n ( the four replaces the n 3(4) = 12 =======
You need an equation for the nth term of the sequence, or some other means of identifying the sequence. In general, they will be a+n, a+2n, a+3n and a+4n although some go for a, a+n, a+2n and a+3n.