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.

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 $a_1$.
