JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

\left(1+x-3 x^2\right)^{2145}=a_0+a_1 x+a_2 x^2+\ldots

, then

a_0-a_1+a_2-a_3+\ldots

ends with___.

View answer and explanation

Correct option: A

Option A (3) is correct because substituting x = -1 into the given expression gives (-3)²¹⁴⁵, and the last digit of this number is 3.

More Questions from BINOMIAL THEOREM

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\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