99Th power of 99+98th power of 99-97th power of 99+96th power of 99--99th power of 3+99th power of 2-99th power of 1 equals?

99Th power of 99+98th power of 99-97th power of 99+96th power of 99--99th power of 3+99th power of 2-99th power of 1 equals?

99^99+(99^98-99^97)+(99^96-99^95)+…+(99^4-99^3)+(99^2-99^1)
=99^99+98×99^97+98×99^95+…+98×99^3+98×99
=99^99+98(99^97+99^95+…+99^3+99^1)
=99^99+98×[(1-99^98)/(1-99^2)](Sum of sequences with a common ratio of 99^2)
=99^99+(99^99-99)÷100

Minus one-third of the power of 1999 multiplied by of the power of 2000 Minus one-third of the power of 1999 by three of the power of 2000

Minus one-third of the power of 1999 multiplied by three equals minus one
Multiply by 3, and it's minus 3.
So answer is:-3