1-x-x2n-r-02 ar x r then a0-a4-a2 n-AB- 3n-1-3 n -1-3 n -1C- 3n-1-2D- 3n-1-2

The correct option is A 3_n+1/2Sum = f(1)+f( 1)/2=(1+1+1)^n (1 1+1)^n/2=3^n+1/2 ...

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Greatest Binomial Coefficients

1+x+x2n=∑r=02...

Question

${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2} a_{r} x^{r} t h e n a_{0} + a_{4} \dots + a_{2 n}$ =

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

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

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

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

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Solution

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Similar questions

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

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

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

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

1, z_{1}, z_{2}, z_{3}, \dots \dots, z_{n - 1}

\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \dots \dots + \frac{1}{3 - z_{n - 1}}

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

E_{1} = a_{0} + a_{3} + a_{6} + . . .

E_{2} = a_{1} + a_{4} + a_{7} + . . .

E_{3} = a_{2} + a_{5} + a_{8} + . . .

(1 + x + x^{2}) n = a 0 + a^{1} x + a^{2} x + a^{3} x^{3} + . . . . . . . a (2^{n}) x (2^{n})

(a^{0} + a^{3} + a^{6} + . . . . . . . . .) = (a^{1} + a^{4} + a^{7}) = (a^{2} + a^{5} + a^{8}) = 3^{(} n - 1),

n \sum r = 1 T_{r} = (3^{n} - 1),

n \sum r = 1 \frac{1}{T_{r}}

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