If $$ \begin{equation} (1+x+x^2+...+x^p)^n=a_0+a_1x+a_2x^2+...+a_{np}x^{np} \end{equation}\label{given}\tag{1} $$
Prove that $$ \begin{equation} S=a_1+2a_2+3a_3+...+np.a_{np}=\frac{1}{2}np(1+p)^n \end{equation}\label{to_prove}\tag{2} $$
Note: I found the way to solve this using calculus. But I need to solve this without using calculus.
Below is the progress I've been able to make:
Substituting $x=1$, $$\underbrace{(1+1+1+\cdots+1)^n}_\textrm{(p+1) terms}=a_0+a_1+a_2+\cdots+a_{np}$$ $$ \begin{equation} \therefore \text{ } (1+p)^n=\sum\limits_{k=0}^{np}a_k \end{equation}\label{res1}\tag{3} $$
We know, $$(1-x)^{-2}=1+2x+3x^2+\cdots+(r+1)x^r+\cdots$$ $$ \begin{equation} \implies\left(1-\frac{1}{x}\right)^{-2}=1+\frac{2}{x}+\frac{3}{x^2}+\cdots+\frac{r+1}{x^r}+\cdots \end{equation}\label{res2}\tag{4} $$
Multiplying $\eqref{given}$ and $\eqref{res2}$, the co-efficient of $x^0$ on RHS would be $$a_0+2a_1+3a_2+\cdots+(np+1)a_{np}$$ $$=(a_0+a_1+a_2+\cdots+a_{np})+(a_1+2a_2+3a_3+\cdots+npa_{np})$$ $$ \begin{equation} =(1+p)^n+S \end{equation}\label{res3}\tag{5} $$
The general term in the expansion of $$\begin{equation}\begin{aligned} (1+x+x^2+...+x^p)^n\times\left(1-\frac{1}{x}\right)^{-2} &=\left(\frac{1-x^{p+1}}{1-x}\right)^n\times\left(1-\frac{1}{x}\right)^{-2}\\ &=x^2(1-x^{p+1})^n(1-x)^{-(2+n)} \end{aligned}\end{equation}\tag{6}\label{res}$$
If you haven't noticed where I'm going at this point, this is what I'm trying to do: to equate the co-efficients of $x^0$ on LHS and RHS of multiplying $\eqref{given}$ and $\eqref{res2}$ with the hope of calculating $S$.
But the problem here is,
- The binomial expansion of $\left(1-\frac{1}{x}\right)^{-2}$ is valid only for $x>1$
- From $\eqref{res}$, it is clear that LHS of $\eqref{given}\times\eqref{res2}$ has no term independent of $x$ and its powers.
Help please!!
Clearly, either combinatorially interpret $a_i$ or through formulas for multinomial expansion, that $a_i=a_{pn-i}$. Therefore
$$S=a_1+2a_2+...+pna_{pn}=\frac{1}{2}pn(a_1+\cdots+a_{pn})=\frac{1}{2}pn(p+1)^n$$