There's a couple ways this can be done. The most straightforward approach is to let #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#.
