Find the sum of each infinite geometric series, if it exists

The first series diverges because each successive term is getting multiplied by -3, so the common ratio is larger than 1 in magnitude.

Assuming the second series is

3 - 3/2 + 3/4 - 3/8 + ...

then it converges because the common ratio between terms is -1/2, which is less than 1 in magnitude.

To find the value of the sum, let S denote the n-th partial sum, i.e. the sum of the first n terms in the series:

S = 3 - 3/2 + 3/4 - 3/8 + ... + 3(-1/2)^(n - 2) + 3(-1/2)^(n - 1)

Multiply both sides by -1/2:

(-1/2) S = -3/2 + 3/4 - 3/8 + 3/16 - ... + 3(-1/2)^(n - 1) + 3(-1/2)^n

When we subtract S and (-1/2) S, we see all the middle terms canceling, and we can solve for S:

S - (-1/2) S = 3 - 3(-1/2)^n

S/2 = 3[1 - (-1/2)^n]

S = 6[1 - (-1/2)^n]

As n gets larger, the (-1/2)^n term will converge to 0, leaving us with

lim (n -> infinity) S = 6

so that the series converges to 6.

it should be a and d. lmk if i'm wrong xd