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If  \left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots \ldots+a_{2 n} x^{2 n}, then  a_0+a_2+\ldots \ldots+a_{2 n} is equal to

 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

We have

\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots \ldots .+a_{2 n} x^{2 n}

Putting x=1 and -1 , we get

1=a_0+a_1+a_2+a_3+\ldots \ldots+a_{2 n}

and  3^n=a_0-a_1+a_2-a_3+\ldots \ldots+a_{2 n}

Adding, we get

1+3^n=2\left(a_0+a_2+a_4+\ldots \ldots+a_{2 n}\right) \\

a_0+a_2+a_4+\ldots . .+a_{2 n}=\frac{3^n+1}{2} .

Posted by

avinash.dongre

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