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consider the infinite geometric series (image attached below) in this image, the lower limit of the summation notation is "n = 1".
a. write the first four terms of the series.
b. does the series diverge or converge?
c. if the series has a sum, find the sum.

a. T1 = -4 , T2 = -4/3 , T3 = -4/9 , T4 = -4/27

b. It is converge

c. The sum to ∞ = -6

Step-by-step explanation:

a.

∵ Tn = -4(1/3)^(n-1)

∵ The lower n = 1

∵ The geometric series Tn = a(r)^(n-1)

∴ T1 = a , T2 = ar , T3 = ar² , T4 = ar³

∴ a = -4 , r = 1/3

∴ T1 = -4

∴ T2 = -4(1/3) = -4/3

∴ T3 = -4(1/3)² = -4/9

∴ T4 = -4(1/3)³ = -4/27

b.

r = 1/3

∴ -1 < r < 1

∴ It is converge because the value of 1/3 when n is a very large

  number will approach to zero

c.

∵ The sum of the geometric series = a(1-(r)^n)/1-r

∵ r^n ≅ 0 when n is a very large number

∴ The sum to ∞ = a/1 - r = -4/(1 - 1/3) = -4/(2/3) = -6

yeah and

step-by-step explanation:

you know what it isnt great cheating

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1