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You can use the identity of y = 1 / (1-x) which is the summation of x^n from n = 0 to infinity. Simply plugin the new values for x (which is 3x) and then multiply the identity by two to find the series for the equation. It would be
? 2(3x)^n from n = 0 to infinity.
2 [ 1 + 3x + 9x^2 + 27x^3 + .... 3^n * x^n ]
The interval of convergence for the identity is (-1, 1). If we plugin our new value of x we'd get
-1 < 3x < 1
Which leaves us with
-1/3 < x < 1/3
So the interval of convergence is (-1/3, 1/3) and the series that represents the function is ? 2(3x)^n from n = 0 to infinity.
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