Question

HELP DUE SOON SOON SOON SOOON

Answer 1

Answer:

second choice

Step-by-step explanation:

The given sums ∑ have j = 1, j = 2, j = 3, j = 4, j = 5

Consider first j = 1 and substitute in the answers to see if we get the first term of the series -3

$-3(\frac{1}{3} )^{1-1} = -3(\frac{1}{3} )^{0}= -3*1= -3$ ✅

2·1 - 5 = 2-5 = -3 ✅

3·1 - 6 = 3-6 = -3 ✅

$5^{1-1} = 5^{0} = 1$ ❌

We see that last choice gives us as first term to be 1 not -3 so eliminate that possibility.

Consider now j = 2 and substitute in the answers to see if we get the second term of the series -1

$-3(\frac{1}{3} )^{2-1} = -3(\frac{1}{3} )^{1}= \frac{-3}{3} = -1$ ✅

2·2 - 5 = 4-5 = -1 ✅

3·2 - 6 = 6-6 = 0 ❌

We see that third choice gives us as second term to be 0 not -1 so eliminate that possibility.

Consider now j = 3 and substitute in the answers to see if we get the third term of the series 1

$-3(\frac{1}{3} )^{3-1} = -3(\frac{1}{3} )^{2}= \frac{-3}{9} = \frac{-1}{3}$ ❌

2·3 - 5 = 6-5 = 1 ✅

We see that first choice gives us as third term to be -1/3 not 1 so eliminate that possibility.

Now we are left with the second choice as the only possibility.