Obtain the following formula without attempting to justify the steps used in the process.
![\[ \sum_{n=0}^{\infty} \frac{(n+1)(n+2)(n+3)}{3!} x^n = \frac{1}{(1-x)^4}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-233d7773b361b45fd982b30988446e1d_l3.png "Rendered by QuickLaTeX.com")
From the previous exercise (Section 10.7, Exercise #18) we have
![\[ \frac{1}{2} \sum_{n=0}^{\infty} (n+1)(n+2) x^n = \frac{1}{(1-x)^3}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2217e49ff853f1707a89dc5e32d7e03a_l3.png "Rendered by QuickLaTeX.com")
Differentiating both sides we then have
![\begin{align*} && \frac{1}{2} \sum_{n=0}^{\infty} (n+1)(n+2)x^n &= \frac{1}{(1-x)^3} \\[9pt] \implies && \frac{1}{2} \sum_{n=0}^{\infty} n(n+1)(n+2) x^{n-1} &= \frac{3}{(1-x)^4} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{(n+1)(n+2)(n+3)}{6} x^n &= \frac{1}{(1-x)^4} \\[9pt] \implies && \sum_{n=0}^{\infty} \frac{(n+1)(n+2)(n+3)}{3!} x^n &= \frac{1}{(1-x)^4}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5f7661419aec89282e6403172c224bd1_l3.png "Rendered by QuickLaTeX.com")
Obtain the following formula without attempting to justify the steps used in the process.
From the previous exercise (Section 10.7, Exercise #18) we have
Differentiating both sides we then have