In the sequence 0, 1, 1, 3, 6, 9, 27, ..., the first term is 0. Subsequent terms are produced by alternately adding and multiplying by each successive integer beginning with 1. For instance, the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on. What is the value of the first term that is greater than 125

645

Step-by-step explanation:

Let's first try and write our series up to 125 before we know the next number.

Well continue from the given series.

0, 1, 1, 3, 6, 9, 27

The next number= 27+4= 31 , then 31 *4 = 124

The next after 124 = 124+5= 129, then 129*5 = 645

So actually the question was supposed to be the term greater than 124 because 125 does not exist in the series.

The number is 645

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

answer: do we write it

step-by-step explanation: