(i recognize that the number is increased by 3 each time, however i put on"t recognize exactly how to discover the umpteenth term:/)

As well as for: 1, 3, 15, 61, 213

The "nth" term can be anything, relying on n"s worth. So we require to compose a formula legitimate for any type of worth of n.

Allow"s see ... we understand the first term, 3, as well as the second is 9. You"ve kept in mind that each succeeding term is 3 times more than the one previously. Because we begin with 3, every term can be composed as a variety of fours:

First term: 3 = 3

Second term: 9 = 3 * 3

Third term: 27 = 3 * 3 * 3

Fourth term: 81 = 3 * 3 * 3 * 3

You"ll note that for each and every term, the variety of 3s increased with each other amounts to the ordinal placement of the term. So for n=4, we require to increase 4 3s with each other. This can be composed utilizing a base and also backer to stand for the variety of 3s we are increasing.

For each and every term n, the solution is 3n, or 3 elevated to the n-th power. The n-th term is figured out by increasing n 3s with each other.

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