Hi can someone please explain how to do this please? 1. Find the general formula that gives the nth term for each of the sequences below
(remember this is NOT the recursive rule).
(a) Odd numbers:
(b) Triangular numbers:
(c) Square numbers:

9514 1404 393

  (a)  an = 2n -1

  (b)  an = n(n +1)/2

  (c)  an = n^2

Step-by-step explanation:

(a) The odd numbers are an arithmetic sequence with first term 1 and common difference 2. The general term is ...

  an = a1 +d(n -1)

  an = 1 + 2(n -1)

  an = 2n -1

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(b) The triangular numbers are a quadratic sequence with first term 1, first first difference 2, and second differences 1.

The first few terms and differences are ...

  sequence: 1, 3, 6, 10, 15, 21, 28, ... (a1 = 1)

  first diff: . . . 2, 3, 4, 5, 6, 7, ... (d1 = 2)

  2nd diff: . . . . 1, 1, 1, 1, 1, ... (d2 = 1)

There is a sort of general formula* that can make use of these numbers to give the general term of the sequence:

  an = a1 +(n -1)(d1 +(n -2)/2(d2))

For the values we have, this gives our general term as ...

  an = 1 + (n -1)(2 +(n -2)/2(1))

  an = n(n +1)/2

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(c) The general term in the sequence of square numbers is simply the square of the term number:

  an = n²

* There are a number of other ways to get to the general formula for triangle numbers. You can search for it, since it is well-known. You can use a quadratic regression tool on the sequence of numbers. You can develop a system of equations to find the coefficients of the quadratic equation.

The general formula given can be used to find polynomial sequences of any degree. Coefficients at successive layers in the formula are of the form (n -k)/k. The first term in parentheses following that is the first k-th difference.

answer: the common difference is -0.10