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Question: If \((1 - x + x^{2})^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ...... + a_{2n}x^{2n}\). Then \(a_{0} + a_{...

If (1−x+x2)n=a0+a1x+a2x2+......+a2nx2n(1 - x + x^{2})^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ...... + a_{2n}x^{2n}. Then a0+a2+a4+......+a2na_{0} + a_{2} + a_{4} + ...... + a_{2n}=

A

3n+12\frac{3^{n} + 1}{2}

B

3n−12\frac{3^{n} - 1}{2}

C

1−3n2\frac{1 - 3^{n}}{2}

D

3n+123^{n} + \frac{1}{2}

Answer

3n+12\frac{3^{n} + 1}{2}

Explanation

Solution

(1−x+x2)n=a0+a1x+a2x2+......+a2nx2n(1 - x + x^{2})^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ...... + a_{2n}x^{2n}

Putting x=1,x = 1, we get (1−1+1)n=a0+a1+a2+....+a2n(1 - 1 + 1)^{n} = a_{0} + a_{1} + a_{2} + .... + a_{2n};

1=a0+a1+a2+.....+a2n1 = a_{0} + a_{1} + a_{2} + ..... + a_{2n} .....(i)

Again putting x=−1x = - 1,we get 3n=a0−a1+a2−......+a2n3^{n} = a_{0} - a_{1} + a_{2} - ...... + a_{2n}......(ii)

Adding (i) and (ii), we get,

3n+1=2[a0+a2+a4+......+a2n]3^{n} + 1 = 2\lbrack a_{0} + a_{2} + a_{4} + ...... + a_{2n}\rbrack

3n+12=a0+a2+a4+.......+a2n\frac{3^{n} + 1}{2} = a_{0} + a_{2} + a_{4} + ....... + a_{2n}