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Addition Rule of Differentiation

If 1+x+ x 2...

Question

If ${(1 + x + x^{2})}^{2 n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{4 n} x^{4 n}$ , then

A

a_{0} + a_{2} + a_{4} + . . . . . . + a_{2 n} = \frac{1}{4} (9^{n} - 1 + {2 a}_{2 n})

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B

a_{0} + a_{2} + a_{4} + . . . . . . + a_{2 n} = \frac{1}{4} (9^{n} + 1 + {2 a}_{2 n})

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C

a_{0} + a_{2} + a_{4} + . . . . . . + a_{2 n} = \frac{1}{4} (9^{n} - 1 - {2 a}_{2 n})

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D

a_{0} + a_{2} + a_{4} + . . . . . . + a_{2 n} = \frac{1}{4} (9^{n} + 1 - {2 a}_{2 n})

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Solution

The correct option is A $a_{0} + a_{2} + a_{4} + . . . . . . + a_{2 n} = \frac{1}{4} (9^{n} + 1 + {2 a}_{2 n})$
$(1 + x + x^{2})^{2 n} = a_{0} + a_{1} (x) + a_{2} x^{2} + . . . a_{4 n} x^{4 n}$
For $x = 1$ , we get
$9^{n} = a_{0} + a_{1} + a_{2} + . . . a_{4 n}$ ... $(i)$
For $x = - 1$ we get
$1 = a_{0} - a_{1} + a_{2} - . . . a_{4 n}$ ... $(i i)$
Adding, $(i)$ and $(i i)$ , we get
$9^{n} + 1 = 2 [a_{0} + a_{2} + a_{4} + . . . a_{4 n}]$

$\Rightarrow \frac{9^{n} + 1}{2} = a_{0} + a_{2} + a_{4} + . . . a_{4 n}$
And $a_{r} = a_{4 n - r}$
Now the middle term will be $a_{2 n}$
Hence
$\frac{1}{2} [9^{n} + 1] = 2 [a_{0} + a_{2} + . . . a_{2 n - 1}] + a_{2 n}$

$\Rightarrow \frac{1}{2} [9^{n} + 1] = 2 [a_{0} + a_{2} + . . . a_{2 n}] - a_{2 n}$

$\Rightarrow \frac{1}{2} [9^{n} + 1] = 2 [a_{0} + a_{2} + . . . a_{2 n}] - \frac{2 a_{2 n}}{2}$

$\Rightarrow \frac{1}{2} [9^{n} + 1 + 2 a_{2 n}] = 2 [a_{0} + a_{2} + . . . a_{2 n}]$

$\Rightarrow \frac{1}{4} [9^{n} + 1 + 2 a_{2 n}] = a_{0} + a_{2} + . . . a_{2 n}$

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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_{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}

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

, find the value of

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

Q. If (1 – x + x²)ⁿ = a₀ + a₁ x + a₂ x² +...+a_2n x²ⁿ, then a₀ + a₂ + a₄ +...+ a_2n equals

(a)

\frac{3^{n} + 1}{2}

\frac{3^{n} - 1}{2}

\frac{1 - 3^{n}}{2}

3^{n} + \frac{1}{2}

{(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} + . . . =

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