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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