Question

初一数学题 急急急急急急！！！！！！！！！！！要具体步骤

3（2的平方+1）（2的四次方+1）(2的八次方+1）（2的一百二十八次方+1）+1
=

Answer 1

3=2²-1
所以原式=(2²-1)(2²+1)(2^4+1)……(2^128+1)+1
=(2^4-1)(2^4+1)……(2^128+1)+1
=(2^8-1)(2^8+1)……(2^128+1)+1
反复用平方差
=2^256-1+1
=2^256

Answer 2

原式
=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=1×(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=(2^4-1)(2^4+1)(2^8+1)……(2^128+1)+1
反复运用平方差公式
=2^256-1+1
=2^256

Answer 3

3（2的平方+1）（2的四次方+1）(2的八次方+1）（2的一百二十八次方+1）+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)...(2^128+1)+1
=(2^4-1)(2^4+1)(2^8+1)...(2^128+1)+1
=(2^8-1)(2^8+1)...(2^128+1)+1
=(2^16-1)(2^16+1)...(2^128+1)+1
=2^256-1+1
=2^256

Answer 4

=(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=1×(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)……(2^128+1)+1
=(2^4-1)(2^4+1)(2^8+1)……(2^128+1)+1
=2^256-1+1
=2^256