Question

Question #7a6ca

Answer 1

`x^2ln (1+x^3) = sum_(n=1)^oo (-1)^(n-1) (x^(3n+2))/n`

Explanation:

Take the MacLaurin series for `ln(1+t)`, which is well known:

`ln (1+t) = sum_(n=1)^oo (-1)^(n-1) t^n/n`

Substitute: `t=x^3`

`ln (1+x^3) = sum_(n=1)^oo (-1)^(n-1) (x^3)^n/n = sum_(n=1)^oo (-1)^(n-1) (x^(3n))/n`

Now multiply by `x^2` term by term:

`x^2ln (1+x^3) = sum_(n=1)^oo (-1)^(n-1) (x^2 x^(3n))/n = sum_(n=1)^oo (-1)^(n-1) (x^(3n+2))/n`

Answer 2

`x^2ln(1+x^3) = -sum_(n=1)^(oo) (-1)^(n)x^(3n + 2)/(n)`

`= x^5 - x^8/2 + x^11/3 - x^14/4 + . . .`

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Recall that

`sum_(n=0)^(oo) t^n = 1/(1-t)`,

which is the derivative of `-ln(1-t)`. So, substitute `t = -x^3` to get

`sum_(n=0)^(oo) (-x^3)^n = 1/(1 + x^3)`,

(whose derivative has no sign change), so that

`intsum_(n=0)^(oo) (-1)^(n)(x^3)^(n)dx = sum_(n=0)^(oo) int(-1)^(n)(x^3)^ndx`

`= sum_(n=0)^(oo) [(-1)^(n)1/(n+1)(x^3)^(n+1)] + C`

`= sum_(n=1)^(oo) [(-1)^(n-1)1/(n)x^(3n)] + C`

`= -sum_(n=1)^(oo) [(-1)^(n)1/(n)x^(3n)] + C`

`= ln(1 + x^3)`, `x >= 0`,

where `C = ln(1 + 0^3) = 0`.

Since the sum depends only on `n`, we can multiply through by `x^2` to get the result:

`color(blue)(x^2ln(1+x^3) = -sum_(n=1)^(oo) (-1)^(n)x^(3n + 2)/(n))`

`= color(blue)(x^5 - x^8/2 + x^11/3 - x^14/4 + . . . )`

whereas that for `ln(1+x^3)`, if you wish to compare, was:

`ln(1+x^3) = x^3 - x^6/2 + x^9/3 - x^12/4 + . . .`

which indeed was just multiplied by `x^2` to get to `x^2ln(1+x^3)`.