Let ${\left( {1 + x} \right)^n} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_n}{x^n}$. If ${a_1},\;{a_2}\,{\rm{and\;}}{a_3}$ are in AP, then the value of $n$ is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

Let

{\left( {1 + x} \right)^n} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_n}{x^n}

. If

{a_1},\;{a_2}\,\mathrm{and\;}{a_3}

are in AP, then the value of

n

View answer and explanation

Correct option: D

Option D (7) is correct because by equating the given condition that a1, a2, and a3 are in AP, we can solve for n, which results in n=7.

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