Multinomial expansion sum 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.
 A: Let's start from here:

$$1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots)$$

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.
A: An easier way to handle this is that since $(1-x)(-x-2) = x^2 + x - 2$, you get
$$\begin{equation}\begin{aligned}
\frac{x^2 + x + 1}{1 - x} & = \frac{x^2 + x - 2 + 3}{1 - x} \\
& = -x - 2 + \frac{3}{1 - x} \\
& = -x - 2 + 3(1 + x + x^2 + x^3 + \ldots) \\
& = 1 + 2x + 3x^2 + 3x^3 + 3x^4 + \ldots \\
& = 1 + 2x + \sum_{i=2}^{\infty}3x^i
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, you have
$$\sum^{50}_{r=1}a_r = 2 + 3(49) = 149 \tag{2}\label{eq2A}$$
As for your error, note you wrote that

Hence, sum
  $ = 3(a_0+ a_1+a_2+\cdots+a_{48}) + 2a_{49}+a_{50} $

You seem to have the indices mixed around, since the coefficient of $2$ should be for $a_1$, not $a_{49}$, and the coefficient of $1$ should be for $a_0$ instead of for $a_{50}$, along with including extra coefficients. Also, as stated in Eevee Trainer's answer, you should not include $a_0$ as the sum starts at $a_1$.
A: A bit late answer but I thought, since you tagged your question also with "generating function", I add a typical generating-function approach using


*

*$(\star)$: $\frac 1{(1-x)^2} = \sum_{n\geq 0}(n+1)x^n$ which can easily be derived by differentiating $\frac 1{1-x}$.


Now, you have
$$\frac{x^2+x+1}{1-x}= \frac{1-x^3}{(1-x)^2}\stackrel{(\star)}{=}(1-x^3)\sum_{n\geq 0}(n+1)x^n$$
$$= \sum_{n\geq 0}(n+1)x^n -\sum_{n\geq 0}(n+1)x^{\color{blue}{n+3}}$$
$$= \sum_{n\geq 0}(n+1)x^n -\sum_{n\geq \color{blue}{3}}(n-2)x^{\color{blue}{n}}\tag{$\star\star$}$$
$$= 1+2x+3x^2 + \sum_{n\geq 3}3x^{n}$$
Using $(\star\star)$. for example, you get
$$\sum_{n=1}^{50}a_r = \sum_{n=1}^{50}(n+1) - \sum_{n=3}^{50}(n-2)$$
$$=\sum_{\color{blue}{n=2}}^{\color{blue}{51}}n - \sum_{\color{blue}{n=1}}^{\color{blue}{48}}n = 49+50+51 -1 = 149$$
