First slide

Binomial theorem for positive integral Index

Question

If for z as real or complex, ${(1 + z^{2} + z^{4})}^{8} = C_{0} + C_{1} z^{2} + C_{2} z^{4} + \dots + C_{16} z^{32}$ ,then

Moderate

A

$C_{0} - C_{1} + C_{2} - C_{3} + \dots + C_{16} = 1$
B

$C_{0} + C_{3} + C_{6} + C_{9} + C_{12} + C_{15} = 3^{7}$
C

$C_{2} + C_{5} + C_{8} + C_{11} + C_{14} = 3^{6}$
D

$C_{1} + C_{4} + C_{7} + C_{10} + C_{13} + C_{16} = 3^{7}$

Solution

${(1 + z^{2} + z^{4})}^{8} = C_{0} + C_{1} z^{2} + C_{2} z^{4} + \dots + C_{16} z^{32} (1)$
Putting z = i, where $i = \sqrt{- 1}$ ,
$(1 - 1 + 1)^{8} = C_{0} - C_{1} + C_{2} - C_{3} + \dots + C_{16}$
or    $C_{0} - C_{1} + C_{2} - C_{3} + \dots + C_{16} = 1$
Also, putting $z = ω,$
${(1 + ω^{2} + ω^{4})}^{8} = C_{0} + C_{1} ω^{2} + C_{2} ω^{4} + \dots + C_{16} ω^{32}$
or $C_{0} + C_{1} ω^{2} + C_{2} ω + C_{3} + \dots + C_{16} ω^{2} = 0 (2)$
Putting $x = ω^{2}$ ,
${(1 + ω^{4} + ω^{8})}^{8} = C_{0} + C_{1} ω^{4} + C_{2} ω^{8} + \dots + C_{16} ω^{64}$
or $C_{0} + C_{1} ω + C_{2} ω^{2} + \dots + C_{16} ω = 0 (3)$
Putting x = 1
   $3^{8} = C_{0} + C_{1} + C_{2} + \dots + C_{16} (4)$
Adding (2), (3) and (4), we have
   $3 (C_{0} + C_{3} + \dots + C_{15}) = 3^{8}$
or $C_{0} + C_{3} + \dots + C_{15} = 3^{7}$
Similarly, first multiplying (1) by z and then putting 1, $ω, ω^{2}$ and adding, we get
$C_{1} + C_{4} + C_{7} + C_{10} + C_{13} + C_{16} = 3^{7}$
Multiplying (1) z²and then putting 1, $ω, ω^{2}$ and adding,
We get $C_{2} + C_{5} + C_{8} + C_{11} + C_{14} = 3^{7}$

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