There's a couple ways this can be done. The most straightforward approach is to let f(x)=3/((1-5x)^2)f(x)=3(1−5x)2 and take derivatives to find it. You can check that f'(x)=30/((1-5x)^3), f''(x)=450/((1-5x)^4), f'''(x)=9000/((1-5x)^5), f''''(x)=225000/((1-5x)^6), etc...
Then f(0)=3, f'(0)=30, f''(0)=450, f'''(0)=9000, f''''(0)=225000, etc...
The Taylor series formula is f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+cdots. Since 450/(2!)=225, 9000/(3!)=1500, and 225000/(4!)=9375, we get the same answer as above: 3+30x+225x^2+1500x^3+9375x^4+cdots.
You can try the ratio test to confirm the radius of convergence is 1/5, but you'd have to find a formula for the coefficients of the powers of x to do so.
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You can do that if you want. I'm going to use a more creative approach. Notice that int 3/((1-5x)^2)\ dx=(3/5)/(1-5x)+C so that d/dx((3/5)/(1-5x))=3/((1-5x)^2). The Taylor series for (3/5)/(1-5x) can be found by using the formula for the sum of a geometric series: a+ar+ar^2+ar^3+cdots=a/(1-r) for |r|<1.
In other words, (3/5)/(1-5x)=3/5+3x+15x^2+75x^3+375x^4+1875x^5+cdots for |5x|<1 leftrightarrow |x|<1/5, making the radius of convergence 1/5.
Now differentiate the preceding series to find the final answer: 3+30x+225x^2+1500x^3+9375x^4+cdots. It's a theorem that guarantees that the radius of convergence for the differentiated series is the same as the original, and is therefore 1/5.
