SOLUTION: write the following in set-builder notation. do not use words. 1.a) {3,7,11,15,...407} 1.b) {1,9,25,49,81,121,...} i figured the pattern for 1.a would be +4 while b would

Question 1139094: write the following in set-builder notation. do not use words.
1.a) {3,7,11,15,...407}
1.b) {1,9,25,49,81,121,...}
i figured the pattern for 1.a would be +4 while b would be x^2.
thanks for helping!!
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

1.a) {3,7,11,15,...407}
difference is
pattern:

since last term is , find total number of terms

{ | ; }
1.b) {1,9,25,49,81,121,...}
using differences you will come up with formula:
1,......9,......25,......49,......81,......121
,...9.....,..16.....24,......32,......40,......
,.......7.,.......7,......8,......8,......
..............0........1.......0->Since we had to take differences twice before we found a constant row, we guess that the formula for the sequence is a polynomial of degree 2, i.e., a quadratic polynomial.

use given terms, set up system of three equations, calculate ,, and

.....eq.1

.....eq.2

.....eq.3
solve the system, and you will get: ,,
so, pattern is:

there is no last term, so can go to infinity
{ | ; }

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