General term of the infinite series, beginning with k= 0, representing the function x^2/(1 + x^3) would be ?

Question

possible solutions....


a. (x^k+1)/k!

b. x^k+1

c. (-1)^k+1 kx^k

d. (-1)^k x^2k+1

e. (-1)^k x^k

f. (-1)^k x^3k

g. (-1)^k x^3k+1

or none of these?

Answer 1

=x^2(1/(1+x^3))
1/(1+x^3)= 1-x^3+x^6-x^9++++(-1)^k x^3k++ and multitplying by x^2 we get (-1)^k*x^(3k+2)( none of these)

Answer 2

INT[x^2/(1 + x^3)] dx = (1/3)ln(1+x^3)
But the power series for ln(1+x) around x = 0 is x -x^2/2 + x^3/3 -...(-1)^(k+1)*x^k/k +..., so the power series for (1/3)ln(1+x^3) at x = 0 is (1/3)(x^3 - x^6/2 +...+(-1)^(k+1)(x^3k)/k +..., and thus:
x^2/(1+x^3) = d( (1/3)ln(1+x^3)/dx = d( (1/3)(x^3 - x^6/2 +...+(-1)^(k+1)(x^3k)/k +...)/dx = ....(-1)^(k+1)*x^(3k-1) ...and this is just (-1)^k*x^3k ---(g), beginning with k = 0.
Regards
Tonio

Answer 3

x^2/(1 + x^3)
=x^2*1/(1 + x^3)
For |x|<1, this
=x^2*(1-x^3+x^6-x^9+x^12-...)
=x^2-x^5+x^8-x^11+x^14-...

So none of the answers is right.