NIMCET 2009 Question Paper with solutions

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Question : 32 of 120

Marks: +1, -0

If

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

, then the value of

a_{2}+a_{4}+a_{5}+. . .+a_{12}

is

1024
64
32
31

Solution:

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

.....(i)

By putting

x=1

on both sides, we get

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

....(ii)
By putting

x=-1

.on both sides of Eq. (i), we get

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

....(iii)

Adding Eqs. (ii) and (iii) gives

64=2[1+a+a_{4}+. . . .+a_{12}]

⇒ \;\; 1+a_{2}+a_{4}+. . .+a_{12}=32

⇒ \;\; a_{2}+a_{4}+. . .+a_{12}=31

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