Answer :
Answer:
[tex]-3x^2+6x^5-9x^8+12x^{11}+...=\sum_{n=1}^{\infty }\left ( -1 \right )^n\,\,3n\,\,x^{3n-1}[/tex]
Step-by-step explanation:
Given : [tex]1+x+x^2+x^3+x^4+\cdots[/tex]
On putting [tex]-x^3[/tex] in place of x , we get [tex]1+\left ( -x^3 \right )+\left ( -x^3 \right )^2+\left ( -x^3 \right )^3+\left ( -x^3 \right )^4+...[/tex]
On simplifying , we get [tex]1-x^3+x^6-x^9+x^{12}+...[/tex]
On differentiating , we get [tex]\left ( 1-x^3+x^6-x^9+x^{12}+... \right )'=-3x^2+6x^5-9x^8+12x^{11}+...[/tex]
Here ,
[tex]-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}[/tex]
Now , we need to express it using summation notation.
[tex]\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}[/tex]