7.Binomial Theorem
normal

If $(1 + x - 3x^2)^{2145} = a_0 + a_1x + a_2x^2 + .........$ then $a_0 - a_1 + a_2 - a_3 + ..... $ ends with

A

$1$

B

$3$

C

$7$

D

$9$

Solution

Put $x = – 1 ; (- 3)^{2145} = a_0 – a_1 + a_2 – a_3 + …… ; – (3)^{2145}$

$ = – (3^4)^{536} · 3$ $\Rightarrow$ ends no. $3$ 

Standard 11
Mathematics

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