TS ICET 23 May 2019 Shift 1 Paper

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Question : 131 of 200

Marks: +1, -0

(1+x+x^2)^n

=

a_0+a_1x+a_2x^2

+ ... +

a_{2n}x^{2n}

${3^n-1}/2$
${3^n+1}/2$
$3^n$
$3^n -1$

Solution:

(1+x+x^{2})^{2}=a_{0}+a_{1} x+a_{2} x^{2} \;\; a_{2 n} x^{2}

Put

x=1

3^{n}=a_{0}+a_{1}+a_{2}+. . .+a_{2 n}

→ 1
Put

x=-1

1=a_{0}-a_{1}+a_{2}-a_{3}+. . .+a_{2 n}

→ 2
(1) – (2)

=3^{n}-1=2(a_{1}+a_{3}+a_{5}+. . . .+a_{2 n-1})

∴ a_{1}+a_{3}+a_{5}+. . . .+a_{2 n-1}=\;{3^{n}-1}/{2}

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