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Laws of Exponents for Real Numbers

If 1+x+ x 2...

Question

If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . . + a_{2 n} x^{2 n}$ , then

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Solution

${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . . + a_{2 n} x^{2 n}$
Multiplying by x both sides
$x {(1 + x + x^{2})}^{n} = a_{0} x + a_{1} x^{2} + a_{2} x^{3} . . . . + a_{2 n} x^{2 n + 1}$ ...(1)
Differentiating both sides w.r.t x we get
$1 + {(1 + x + x^{2})}^{n} + n x {(1 + x + x^{2})}^{n - 1} (1 + 2 x) = a_{0} + 2 a_{1} x + 3 a_{2} x^{2} . . . . + (2 n + 1) a_{2 n} x^{2 n}$ ...(2)
A) Substituting $x = 1$ in (1) we get
$a_{0} + a_{1} + . . . . + a_{2 n} = 3^{n}$
B) Substituting $x = - 1$ in (1) we get
$a_{0} - a_{1} + a_{2} - . . . . + a_{2 n} = 1$
C) Substituting $x = - 1$ in (2) , we get
$a_{0} - 2 a_{1} + {3 a}_{2} - . . . . + (2 n + 1) a_{2 n} = n + 1$
D) Substituting $x = 1$ in (2) , we get
$a_{0} + 2 a_{1} + {3 a}_{2} + . . . . + (2 n + 1) a_{2 n} = (n + 1) 3^{n}$

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Similar questions

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}

a_{0} + a_{3} + a_{6} + . . . =

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . . . . . + a_{2 n} x^{2 n}

then the value of

a_{0} + a_{3} + a_{6} + . . . . .

n

be a positive integer such that

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}

a_{r} =

(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{2 n} x^{2 n}

a_{0} + a_{2} + a_{4} + . . . . + a_{2 n}

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

, then the value of

a_{0} + a_{2} + a_{4} + \dots + a_{2 n}

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