Is it Summation really?

$Using\quad binomial\quad theorem\quad for\quad ((1-x)^{ -1 }),\quad we\quad get\quad ((x^{ 2 }+x+1)\times (1+x+x^{ 2 }...))\\ Therefore\quad Coefficient\quad are\quad :\quad (a_{ 0 }=1,a_{ 1 }=2,a_{ 2 }=3,a_{ 3 }=3,a_{ 4 }=3,.....).\\ Thus\quad sum\quad becomes\quad 1+2+3+3+...=3\times 48+2+1=147.\\ ANS.$

$\begin{aligned} \frac{x^2+x+1}{1-x} & = - x - 2 + \frac{3}{1-x} \quad \quad \small \color{#3D99F6}{\text{By long division}} \\ & = - 2 - x + 3 \times \color{#3D99F6}{\frac{1}{1-x}} \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = - 2 - x + 3 \color{#3D99F6}{(1 + x + x^2 + x^3 + \cdots)} \\ & = 1 + 2x + 3x^2 + 3x^3 + 3x^4 + \cdots \end{aligned}$

$\implies \displaystyle \sum_{i=0}^{49} a_i = 1 + 2 + 48(3) = \boxed{147}$

---

More about Maclaurin's series .