JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

Let

n

be a positive integer such that

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

, then

a_0+a_1+a_2+\ldots+a_{n-1}

View answer and explanation

Correct option: C

\frac{1}{2}\left(3^n-a_n\right)

Option C (

\frac{1}{2}(3^n-a_n)

) is correct because it utilizes the symmetry of the coefficients and the total sum to isolate the desired partial sum.

More Questions from BINOMIAL THEOREM

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n