First slide

Binomial theorem for positive integral Index

Question

If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . + a_{2 n} x^{2 n}$ , then

Moderate

A

$a_{0} - a_{2} + a_{4} - a_{6} + \dots = 0$ ,if n is odd
B

$a_{1} - a_{3} + a_{5} - a_{7} + \dots = 0$ ,if n is even
C

$a_{0} - a_{2} + a_{4} - a_{6} + \dots . 0, if n = 4 p, p \in I^{+}$
D

$a_{1} - a_{3} + a_{5} - a_{7} + \dots . = 0, if n = 4 p + 1, p \in I^{+}$

Solution

$∵ {(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$
Putting $x = i (i = \sqrt{- 1})$
Then, we get
${(1 + i + i^{2})}^{n} = (a_{0} - a_{2} + a_{4} - a_{6} + \dots) + i (a_{1} - a_{3} + a_{5} - a_{7} + \dots)$
$\Rightarrow i^{n} = (a_{0} - a_{2} + a_{4} - a_{6} + \dots) + i (a_{1} - a_{3} + a_{5} - a_{7} + \dots)$
if n is odd, then Re(iⁿ) = 0
$\Rightarrow a_{0} - a_{2} + a_{4} - a_{6} + \dots = 0$
if n is even, then Im (iⁿ) = 0
$\Rightarrow a_{1} - a_{3} + a_{5} - a_{7} + \dots = 0$

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