come up with a formula for the function f(x) whose expansion in powers of x is ((x^2)/2! )+((x^3)/3! )+((x^4)/4! )+((x^5)/5! )+...+((x^n)/n! )

Question

anonymous · Answer

looks a lot like $e^x$ except you are missing the first two terms

anonymous · Answer

clear or no?

anonymous · Answer

i would say $f(x)=e^x-1-x$

anonymous · Answer

it's not e^x maybe i think it's second one  thankyou

anonymous · Answer

oh it is not $e^x$ because you are missing the first two terms of the expansion

anonymous · Answer

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ so 
$$e^x-1-x=\frac{x^2}{2!}+\frac{x^2}{3!}+...$$

anonymous · Answer

can you help me on another question?

anonymous · Answer

sure go ahead and post, i will help if i can

anonymous · Answer

it's two parametric function to get the volume

anonymous · Answer

probably not, but post in a new question and i bet you will get lots of replies 
i will try it too

anonymous · Answer

i hate integrating, it is a pain