NIMCET 2019 Question Paper with solutions

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Question : 14 of 120

Marks: +1, -0

If

(1 + x — 2x^2)^6

= 1 +

a_1x

+

a_2x^2

+ ... +

a^12x^12

, then the value

a_2 + a_4

+

a_6

+ — +

a_12

is

29
30
31
32

Solution:

We have,

(1+x-2x^{2})^{6}=1+a_{1}x+a_{2}x^{2}+. . .+a_{12}x^{12}

Put

x=1

on both sides we. get

0=1+a_{1}+a_{2}+a_{3}+a_{4}+. . .+a_{12}

...(i)
Put

x=-1

on both sides, we get

64=1-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}...+a_{12}

....(ii)
Adding Eqs. (i) and (ii). we get

64=2+2(a_{2}+a_{4}+a_{6}+. . .a_{12})

⇒a_{2}+a_{4}+a_{6}. . .+a_{12}=31

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