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Question

If (1+x+x2)20=a0+a1x+a2x2++a40x40, then the value of a20a21+a22a219 is 


Correct Answer
A
12a20(1a20)
Your Answer
B
12a20(1+a20)
Your Answer
C
12a220
Your Answer
D
\N

Solution

The correct option is A 12a20(1a20)
Given 
(1+x+x2)20=40r=0arxr   (1)

Now, replacing x by 1x, we get
(1+1x+1x2)20=40r=0ar(1x)r(1+x+x2)20=40r=0arx40r   (2)

Comparing equation (1) and (2), we get
a0=a40a1=a39
So, ar=a40r

Replacing x by 1x in (1), we get
(11x+1x2)20=a0a1x+a2x2+a40x40
(1x+x20)20=a0x40a1x39+a2x38+a40

Clearly,
a20a21+a22+a240
is the coefficient of x40 in (1+x+x2)20 (1x+x2)20
= Coefficient of x40 in (1+x2+x4)20 
=coefficient of y20 in (1+y+y2)20, where y=x2 
= a20.

Hence, a20a21+a22+a240=a20
(a20a21+a22a219)+a220
+(a221+a222a223++a240)=a20
ar=a40r, the above equation can be reduced to
2(a20a21+a22a219)+a220=a20
a20a21+a22a219=a202(1a20)

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