First slide

Binomial theorem for positive integral Index

Question

If ${(1 + x + x^{2})}^{48} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{96} x^{96}$ then the value of $a_{0} - a_{2} + a_{4} - a_{6} + \dots + a_{96}$

Moderate

A

-1
B

0
C

1
D

48

Solution

Putting x = i, we get
$\begin{aligned} {(1 + i + i^{2})}^{48} = a_{0} + a_{1} i + a_{2} i^{2} + \dots + a^{98} i^{96} \\ \Rightarrow i^{48} = (a_{0} - a_{2} + a_{4} - \dots) + i (a_{1} - a_{3} + \dots) \end{aligned}$
Equating the real parts we get
$a_{0} - a_{2} + a_{4} - a_{6} + \dots + a_{96} = 1$

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