Q: What the sequence for n plus 3?

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Put N = 10 and what you get is U10 = 5*10 + 3 = 50 + 3 = 53.

Put n = 1, 2, 3, 4 etc in the expression 5n + 2 and evaluate to get the sequence.

Strangely enough, it is 9n + 1 for n = 1, 2, 3, ...

A(1) = 12A(4) = 3 A(10) = -15.

5+n+3 = 8+n

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Put N = 10 and what you get is U10 = 5*10 + 3 = 50 + 3 = 53.

Put n = 1, 2, 3, 4 etc in the expression 5n + 2 and evaluate to get the sequence.

Strangely enough, it is 9n + 1 for n = 1, 2, 3, ...

A(1) = 12A(4) = 3 A(10) = -15.

The nth term of a sequence is the general formula for a sequence. The nth term of this particular sequence would be n+3. This is because each step in the sequence is plus 3 higher than the previous step.

The sequence is poorly defined. 1+3+5 appears to be a sequence of odd numbers. However, that cannot end in 100: it can attain the values of 99 or 101. Obviously the answer will depend on which one of these is the final number. An alternative is that the sequence is not that of odd numbers but some other sequence: for example, t(n) = (29n3 - 174n2 + 399n - 214)/40 which, for n = 1, 2, 3, generates the sequence 1, 3, 5, 11.35, 26.4, 54.5, 100 whose sum is 201

4 5 6 7 8 9 10 11 12 13

64. It's the sequence f(n) = n^3

5+n+3 = 8+n

7n+10

The answer is any member of the sequence 40n - 22 for n = 1, 2, 3, ...The answer is any member of the sequence 40n - 22 for n = 1, 2, 3, ...The answer is any member of the sequence 40n - 22 for n = 1, 2, 3, ...The answer is any member of the sequence 40n - 22 for n = 1, 2, 3, ...

If the first integer is "n", then the second integer would be "n + 1", and the third, "n + 2". The sum of all this is (n) + (n + 1) + (n + 2) = 3n + 3. In other words, three times the first number in the sequence, plus 3.