I was solving some questions of Multinomial Theorem from Higher Algebra by Hall and Knight when I encountered this question.
$\text{If }\left(1+x+x^{2}+\ldots+x^{p}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n p} x^{n p},\text{ Prove that,}$
$(1) a_{0}+a_{1}+a_{2}+\ldots+a_{n p}=(p+1)^{n}$
$(2) a_{1}+2 a_{2}+3 a_{3}+\ldots+n p \cdot a_{m p}=\frac{1}{2} n p(p+1)^{n}$
Here is the Solution that was given,
$\text{The first part is obtained by putting }x=1,\text{ for then}$
$1+x+x^{2}+\ldots+x^{p}=p+1$
$\text{For the second part, change x into }1+x\text{; thus}$
$a_{0}+a_{1}(1+x)+a_{2}(1+x)^{2}+a_{3}(1+x)^{3}+\ldots+a_{n p}(1+x)^{n p}$
$=\left\{1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots+(1+x)^{p}\right\}^{n}$
$=\{1+1+1+\ldots \text { to }(p+1) \text { terms }+\langle 1+2+3+\ldots \text { to } p \text { terms }) x+\text { higher powers of } x\}^{n}$
$=\left\{p+1+\frac{p(p+1)}{2} x+\text { higher powers of } x\right\}^{n}=(p+1)^{n}\left(1+\frac{p x}{2}+\dots\right)^{n}$
$=(p+1)^{n}\left(1+\frac{n p}{2} x+\text { higher powers of } x\right)$ $\quad$
$\text{Hence by equating the coefficients of }x$
$a_{1}+2 a_{2}+3 a_{3}+\ldots+n p a_{n p}=\frac{1}{2} n p(p+1)^{n}$
I am not able to understand how we have got the last two equations.
Any help would be appreciated.
Consider $$ (a + b)^n = \sum_{i=0}^n {n \choose i} a^{n-i} b^i $$ and let $a = 1$ and $b = \frac{px}{2}$; you get \begin{align} (1 + \frac{px}{2})^n &= \sum_{i=0}^n {n \choose i} 1^{n-i} \left(\frac{px}{2}\right)^i \\ &= \sum_{i=0}^n {n \choose i} \left(\frac{px}{2}\right)^i \\ &= {n \choose 0} \left(\frac{px}{2}\right)^0 + {n \choose 1} \left(\frac{px}{2}\right)^1 + \ldots \\ &= 1 + {n \choose 1} \left(\frac{px}{2}\right)^1 + \ldots \\ &= 1 + n \left(\frac{p}{2}\right) x + \ldots \\ &= 1 + \left(\frac{np}{2}\right) x + \ldots \\ \end{align} where the ellipses indicate terms with higher powers of $\frac{px}{2}$, hence terms in which $x$ appears to powers greater than one.