Question

2的100次方减2的99次方减2的98次方......减2的2次方减2

Answer 1

2^100 - (2^99 + 2^98 + ...... + 2^2 + 2)
= 2^100 - A

A = 2^99 + 2^98 + ...... + 2^2 + 2 ----------------(1)
2A = 2^100 + 2^99 + 2^98 + ...... + 2^3 + 4 -------(2)

(2) - (1)
A = 2^100 - 2

所以原式 ＝ 2

Answer 2

2^99+2^98+2^97+……+2^2+2^1
=2*(1-2^99)/(1-2)
=2^100-2
所以
2^100-2^99-2^98-……-2^2-2
=2^100-(2^99+2^98+……+2^2+2)
=2^100-(2^100-2)
=2

Answer 3

2的100次方等于2的99次方的2倍，2的99次方是2的98次方的2倍……依次减下去，最后不就剩下个2的2次方减2吗？！结果不言而喻2

Answer 4

解:设原式为A
则2A=2^101-2^100-.....-2^3-2^2
2A-A=2^101-2*2^100+2=2^101-2^101+2=2

Answer 5

2