Consider the series Summation from n equals 0 to infinityStartFraction (negative 1 )Superscript n Baseline x Superscript n Over (n plus 1 )exclamation mark EndFraction . (a) Find the series' radius and interval of convergence. (b) For what values of x does the series converge absolutely? (c) For what values of x does the series converge conditionally?

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

1 ?

step-by-step explanation:

u said 1+1+1+1+1+1+1+1+1+1+1+1*0 so ur adding 1 but then multiplying by 0 so its 0 but then u said 1*0+1 so now ur adding 1 so the answer is juan ?

~leah

the first sequence is an arithmetic sequence.

n=2.7

step-by-step explanation: