KVPY SX SB Exam 2015 Question Paper

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Question : 82 of 120

Marks: +1, -0

Let

(1 + x + x^2)^2014 = a_0 + a_1x + a_2x^2 + a_3x^3 + ..... + a_4028x^4028

, and let

A = a_0 - a_3 + a_6 - .... + a_4026,

B = a_1- a_4 + a_7 - ....... -a_4027,

C = a_2 - a_5 + a_8 - .... + a_4028,

Then -

$|A| = |B| > |C|$
$|A| = |B| < |C|$
$|A| = |C| > |B|$
$|A| = |C| < |B|$

Solution:

(1 + x + x^2 )^ 2014 = a_0 + a_1x + a_2x ^2

+ a_3x ^3 + ..... + a_4028x ^4028

put x = – 1

1 = 1 = a_0 - a_1 + a_2- a_3 + a_4-a_5

+ a_6 - a_7 …. \;\;

... (1)
put

x = - ω

(2ω)^ 2014 = (1 - ω + ω^2 ) ^2014 = a_0 - a_1ω

+ a_2ω^2 - a_3 + a_4ω - a_5ω^ 2 + a_6……

...(2)
Put

x =- ω^ 2

(2ω^ 2 )^ 2014 = (1 - ω^ 2 + ω) ^2014 = a_0 - a_1ω^ 2

+ a_2ω - a_3 + a_4ω^ 2 - a_5ω……

...(3)
Now, (1) + (2) + (3)
⇒

1+ (2ω) ^2014+ (2ω^2 )^2014 = 3(a_0- a_3 + a_6… ...)

⇒

a_0 - a_3 + a_6…… ={1 +2 ^2014 ω+2^2014 ω^2}/3

A ={1-2^2014}/3

|A|={ 2^2014-1}/3

and

(1) + (2) × ω + (3)ω^ 2

⇒

{1 +2 ^2014 .ω ^2014.ω+2^2014 .(ω^2) ^2014.ω^2}/3 = a_2 - a_5 + a_8……

⇒

{1 +2 ^2014 +ω ^2015 +2^2014 .(ω ) ^ 4030 }/3 = C

⇒

C={1-2^2014}/3⇒|C|={ 2^2014-1}/3

and similarly

(1) + (2) ×ω^ 2 + (3) × ω

B={1 +2 ^2014 .ω ^2014.ω^2+2^2014 .(ω^2) ^2014.ω }/3

={1 +2 ^2014 .ω ^2016. +2^2014 .(ω ) ^4029. }/3

|B|={1+2^2015}/3

∴

|B| > |A| = |C|

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