If `(1+x-3x^2)^2145 = a_0+a_1 x+a_2 x^2+...` then `a_0-a_1+a_2-..` ends with

Options

A. 1
B. 3
C. 7
D. 9

Answer

b

Explanation

We have,
`( 1 + x - 3x^(2)) ^(2145) = a_(0) + a_(1)x + a_(2)x^(2)+…`
On putting x = - 1 , we get
` a_(0) - a_(1) + a_(2) - …= (-3)^(2145)`
But we know that,
` 3^(1) = 3,3^(2) = 9,3^(3) = 27,3^(4)= 81`
` therefore a_(0) - a_(1) + a_(2) + ...= [(-3)^(4)]^(536) (-3)^(1)`
` therefore ` End digit of ` (-3)^(2145)`
= End digit of `[(-3)^(4) ]^(536) xx " End digit of " (-3)^(1)`
` = 1 xx 3= 3` .