1 - (1/2)(x-3) + (1/4)(x-3)^2 - (1/8)(x-3)^3 + ... + (-1/2)^n(x-3)^n + ... a) Wh

ID: 3346498 • Letter: 1

Question

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

a) What values of x does the series converge?
b) What is the sum?

c)What series do you get if you differentiate the given series term by term?

d)For what values of x does the new series (from part c) converge?

e)What is the new series sum?
f)Integrate the original series term by term. What new series do you get?

g)For what values of x does the new series (from part f) converge?

h) What is the sum of this new series?

Please help I will rate!! Show steps fully please

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

let (x-3)/2 =p

lthen S = 1 -p + p^2 - p^3 + P^4 - ...

S = 1/(1-p).

S1=let nwe series is S1

=integration [S*dx]

=integration [S*2*dp]

= 2*integration [S*dp]

= 2*integration [1/(1-p)*dp]

= -2*ln(1-p)

= -2*ln[ (5-x)/2]

f) coverge when x< 5

h) sum = S1 =-2*ln[ (5-x)/2]

= 2*ln[2/(5-x)]

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