First slide

Binomial theorem for negative integral and rational Index

Question

If $\frac{x^{2} + x + 1}{1 - x} = a_{0} + a_{1} x + a_{2} x^{2} + \dots$ , then $\sum_{r = 1}^{50} a_{r}$ is equal to

Easy

A

148
B

146
C

149
D

150

Solution

$\begin{matrix} \frac{(x^{2} + x + 1) (1 - x)}{(1 - x)^{2}} = (1 - x^{3}) (1 - x)^{- 2} \\ = (1 - x^{3}) (1 + 2 x + 3 x^{2} + \dots . . . + (r + 1) x^{r} + . . . . . . . . . . . .) \end{matrix}$
Now, $a_{r} = (r + 1) - (r - 2) = 3$
But a₁ = 2
So, $\sum_{r = 1}^{50} a_{r} = 2 + 49 \times 3 = 149$

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