Lf 1-x-x2n-a0-a1x-a2x2-a2nx2n then a02-a12-a22-a2n2-

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lf 1+x+x2n=...

Question

lf $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . + a_{2 n} x^{2 n}$ then $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{2 n}^{2} =$

0

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a_{n}

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- a_{n}

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a_{n - 1}

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Solution

The correct option is B $a_{n}$
Let us consider $n = 2$
Therefore $(1 + x + x^{2})^{2}$
$= 1 + x^{2} + x^{4} + 2 x + 2 x^{3} + 2 x^{2}$
$= 1 + 2 x + 3 x^{2} + 2 x^{3} + x^{4}$
Hence $a_{0} = 1$
$a_{1} = 2$
$a_{2} = 3$
$a_{3} = 2$
$a_{4} = 1$
Therefore $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + a_{4}^{2}$
$= 1 - 4 + 9 - 4 + 1$
$= 11 - 8$
$= 3$
$= a_{2}$ which implies $a_{n}$
This is only satisfied by Option B

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Q. If

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a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}

Q. Let

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, such that

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lf (1+x+x2)n=a0+a1x+a2x2+......+a2nx2n then a20−a21+a22−…+a22n=

The correct option is B anLet us consider n=2Therefore (1+x+x2)2=1+x2+x4+2x+2x3+2x2=1+2x+3x2+2x3+x4Hence a0=1a1=2a2=3a3=2a4=1Therefore a20−a21+a22−a23+a24=1−4+9−4+1=11−8=3=a2 which implies anThis is only satisfied by Option B

lf $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . + a_{2 n} x^{2 n}$ then $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{2 n}^{2} =$