You can first use the well-known series for #e^(x)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+\cdots#
Replace #x# with #2x#:
#e^(2x)=1+2x+(2x)^2/(2!)+(2x)^3/(3!)+(2x)^4/(4!)+\cdots#
Then multiply everything by #x# and simplify:
#xe^(2x)=x+2x^2+2^2/(2!)x^3+2^3/(3!)x^4+2^4/(4!)x^5+\cdots#
#=x+2x^2+2x^3+4/3 x^4+16/24 x^5+32/120 x^6+64/720 x^7+128/5040 x^8+\cdots#
#=x+2x^2+2x^3+4/3 x^4+2/3 x^5+4/15 x^6+4/45 x^7+8/315 x^8+\cdots#
You can also do this by using the formula #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+\cdots#, where #f(x)=xe^(2x)#.
Calculating some derivatives here gives #f'(x)=e^(2x)+2xe^(2x)#, #f''(x)=2e^(2x)+2e^(2x)+4xe^(2x)=4e^(2x)+4xe^(2x)#, #f'''(x)=8e^(2x)+4e^(2x)+8xe^(2x)=12e^(2x)+8xe^(2x)#, #f''''(x)=32e^(2x)+16xe^(2x)#, etc...
Hence, #f(0)=0#, #f'(0)=1#, #f''(0)=4#, #f'''(0)=12#, #f''''(0)=32#, etc..., resulting in
#x+4/(2!)x^2+12/(3!)x^3+32/(4!)x^4+\cdots#
#=x+2x^2+2x^3+4/3 x^4+\cdots#
