Question

If 4x8^m=2^5 then 1+m^1+m^2+………..m^2019=

Answer 1

Given:

• 4x8^m=2^5

To Find:

• 1+m^1+m^2+………..m^2019

Solution:

⇒  8^m = 2^5/4

⇒  8^m = 8

⇒  m = 1

___________________________

∴ 1+m^1+m^2+………..m^2019= 1 + [ 1+1+1+1+1+...(2019 times)]

⇒ 1+m^1+m^2+………..m^2019= 1 + 2019

⇒ 1+m^1+m^2+………..m^2019= 2020

____________________________

Final Answer → 2020

Answer 2

Answer:

We know that D=b

2

−4ac

D=4(1+3m)

2

−4(1+m

2

)(1+8m)

=4(1+9m

2

+6m−(1+8m+m

2

+8m

3

))

=4(8m

2

−2m−8m

3

)

=−8(4m

3

−4m

2

+m)

=−8m(4m

2

−4m+1)