Given,
$$ \frac{x^2+x+1}{1-x}= a_0+a_1x +a_2x^2+\cdots $$
then, find the sum:
$$ \sum^{50}_{r=1}a_r $$
I knew using multinomial expansion, that
$$ (1-x)^{-1} = 1+ x+x^2+\cdots$$
Hence, $$ \frac{x^2+x+1}{1-x}=1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots) $$
Hence, sum
$ = 3(a_0+ a_1+a_2+\cdots+a_{48}) + 2a_{49}+a_{50} $
So, sum should be $147+3 = 150$. But, I was wrong. Any hints?
The correct answer was
149
Edit: from @John's answer, I got to know that I used symbol $a_r$ for multinomial coeff. and sum coeff. both. The latter $a_r$ was for multinomial actually.
Best Answer
Let's start from here:
Let us rewrite this in a somewhat more friendly way as well by writing out what the sum would be, in effect:
$$\begin{array}{rrrrr} 1 & +x & +x^2 & +x^3 & \cdots \\ & x & +x^2 & +x^3 & \cdots \\ & & x^2 &+x^3 & \cdots \\ \hline 1 & +2x &+ 3x^2 &+ 3x^3 & \cdots \end{array}$$
From the $x^2$ term onwards, there's always an $x^k$ ($k \ge 2$) expression in each of the infinite sums once you distribute the outside term, so there's three of them.
You want the sum of the coefficients for the $x,x^2,x^3,\cdots,x^{50}$ terms as well. Thus you want $2 + 3\cdot49$ as your sum, which is $149$.
Your error is likely that you just forgot that you're not meant to count $a_0$ (your constant term): the sum is from $r=1$ to $50$ of $a_r$. This would explain why you're precisely $1$ over the correct answer.