$\displaystyle{\frac{18}{(1+3x)^3}}$ $=\sum_{n=0}^\infty n(n-1)(-3)^n x^{n-2}$
If i got up to this, how could i get $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ ?
When i tried to multiply both side, some people says n-m some says n+m for $x^m$
Could someone kindly show me the working out please?
My working out:
$\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ = $=\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$
Multiply (1-2x) on both side I got
= $\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$ - $2\sum_{n=0}^\infty [((n-1)(n-2)(-3)^{(n-1)}))/18] x^{(n-2)}$
$=\sum_{n=0}^\infty [(5n-4)(n-1)(-3)^n /54 ] x^{(n-2)}$
So we are given the (interesting) equality $$ \frac{{18}}{{(1 + 3x)^3 }} = \sum\limits_{n = 2}^\infty {n(n - 1)( - 3)^n x^{n - 2} } , $$ for $x$ in a neighborhood of $0$. Hence, $$ \frac{{18}}{{(1 + 3x)^3 }} = \sum\limits_{n = 0}^\infty {(n + 2)(n + 1)( - 3)^{n + 2} x^n } , $$ and in turn $$ \frac{1}{{(1 + 3x)^3 }} = \sum\limits_{n = 0}^\infty {\frac{{(n + 2)(n + 1)( - 3)^n }}{2}x^n } . $$ Thus, $$ \frac{{ - 2x}}{{(1 + 3x)^3 }} = \sum\limits_{n = 0}^\infty {\frac{{ - 2(n + 2)(n + 1)( - 3)^n }}{2}x^{n + 1} } = \sum\limits_{n = 1}^\infty {\frac{{ - 2(n + 1)n( - 3)^{n - 1} }}{2}x^n } . $$ Therefore, $$ \frac{{1 - 2x}}{{(1 + 3x)^3 }} = 1 + \sum\limits_{n = 1}^\infty {\frac{{( - 3)^n }}{2}x^n \bigg[(n + 2)(n + 1) + \frac{2}{3}(n + 1)n \bigg]}, $$ or $$ \frac{{1 - 2x}}{{(1 + 3x)^3 }} = 1 + \sum\limits_{n = 1}^\infty {\frac{{( - 3)^n }}{2}\bigg[\frac{5}{3}n^2 + \frac{{11n}}{3} + 2\bigg]x^n } $$ (confirmed numerically).