If - n - N -1- x - x 2 n - a 0- a 1 x - a 2 x 2- a 2 n x 2 n where a 1 a 2- - a 2 n - Zand a02-a12-a22-a32- - -a2 n2-k an then k-A- 1B- 2C- 1 - 2D- 0

The correct option is A 1 (1+x+x^2)^n=a_0+a_1x+a_2x^2+......+a^2nx^2n Replacing by 1 x we have ( 1 1 x+1 x^2 )^n=a_0 a_1 x+a_2 x^2+.......+a_2n x^2n Now a_0^2 ...

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Byju's Answer

Standard XI

Mathematics

Inequality

If ∀ n ∈ N ,1...

Question

If $\forall n \in N, (1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . + a_{2 n} x^{2 n} w h e r e a_{1} a_{2} . . . . . a_{2 n} \in Z$
and $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . + a_{2 n}^{2} = k a_{n}$ then k=

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Solution

The correct option is A
1

$(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . + a^{2 n} x^{2 n}$
Replacing by $- \frac{1}{x}$ we have
$(1 - \frac{1}{x} + \frac{1}{x^{2}})^{n} = a_{0} - \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} + . . . . . . . + \frac{a_{2 n}}{x^{2 n}}$
Now $a_{0}^{2} - a_{1}^{2} - a_{3}^{3} + . . . . . . . + a_{2 n}^{2}$ =term independent of x in $(1 + x + x^{2})^{n} (1 - \frac{1}{x} + \frac{1}{x^{2}})^{n}$
=term independent of x in $(1 + x + x^{2})^{n} (\frac{1 - x + x^{2}}{x^{2}})^{n}$
=term independent of x in $\frac{(1 + x^{2} + x^{4})^{n}}{x^{2 n}}$
=term independent of $x^{2 n}$ in $(1 + x^{2} + x^{4})^{n}$
=term independent of $x^{n}$ in $(1 + x + x^{2})^{n}$
$= a_{n}$

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If ∀ n ∈ N,(1+x+x2)n=a0+a1x+a2x2+......+a2nx2n where a1a2.....a2n∈Z and a20−a21+a22−a23+.....+a22n=kan then k=

If $\forall n \in N, (1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . + a_{2 n} x^{2 n} w h e r e a_{1} a_{2} . . . . . a_{2 n} \in Z$
and $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + . . . . . + a_{2 n}^{2} = k a_{n}$ then k=