Question

Start with the geometric power series LaTeX: 1+x+x^2+x^3+x^4+\cdots Substitute LaTeX: -x^3in place of LaTeX: x, and then take a derivative term by term. Express this new series by writing out at least the first five terms, and also express it using summation notation.

Answer 1

Answer:

$-3x^2+6x^5-9x^8+12x^{11}+...=\sum_{n=1}^{\infty }\left ( -1 \right )^n\,\,3n\,\,x^{3n-1}$

Step-by-step explanation:

Given : $1+x+x^2+x^3+x^4+\cdots$

On putting $-x^3$ in place of x , we get $1+\left ( -x^3 \right )+\left ( -x^3 \right )^2+\left ( -x^3 \right )^3+\left ( -x^3 \right )^4+...$

On simplifying , we get $1-x^3+x^6-x^9+x^{12}+...$

On differentiating , we get $\left ( 1-x^3+x^6-x^9+x^{12}+... \right )'=-3x^2+6x^5-9x^8+12x^{11}+...$

Here ,

$-3x^2=\left ( -1 \right )^1\,3(1)x^{3(1)-1}\\6x^5=\left ( -1 \right )^2\,\,3(2)\,\,x^{3(2)-1}\\-9x^8=\left ( -1 \right )^3\,\,3(3)\,\,x^{3(3)-1}\\12x^{11}=\left ( -1 \right )^4\,\,3(4)\,\,x^{3(4)-1}$

Now , we need to express it using summation notation.

$\left ( 1-x^3+x^6-x^9+x^{12}+... \right )'=-3x^2+6x^5-9x^8+12x^{11}+...=\sum_{n=1}^{\infty }\left ( -1 \right )^n\,\,3n\,\,x^{3n-1}$