First slide

Binomial theorem for positive integral Index

Question

If ${(1 + x - 2 x^{2})}^{6} = \sum_{r = 0}^{12} A_{r} x^{r}$ , then value of $A_{2} + A_{4}$ $+ A_{6} + \dots + A_{12}$ is

Moderate

A

30
B

31
C

32
D

none of these

Solution

Clearly $A_{0} = 1$
Putting $x = 1$ and $x = - 1,$ we get
$0 = A_{0} + A_{1} + A_{2} + \dots + A_{12}$
and $(- 2)^{6} = A_{0} - A_{1} + A_{2} + \dots + A_{12}$
Adding we get
$\begin{aligned} 64 = 2 ⌊A_{0} + A_{2} + \dots + A_{12}⌋ \\ \Rightarrow A_{2} + A_{4} + \dots + A_{12} = 31 \end{aligned}$

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