代数计算(a+b+c)²（2+1）（2²+1）（2的四次方+1）（2的八次方+1）······（2的三十二次方+1）+1
2019-05-07
代数计算
(a+b+c)²
（2+1）（2²+1）（2的四次方+1）（2的八次方+1）······（2的三十二次方+1）+1
优质解答
1、
原式=[(a+b)+c]²
=(a+b)²+2(a+b)c+c²
=a²+b²+c²+2ab+2bc+2ca
2、
原式=1*(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2²-1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^16-1)(2^16+1)(2^32+1)+1
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64
1、
原式=[(a+b)+c]²
=(a+b)²+2(a+b)c+c²
=a²+b²+c²+2ab+2bc+2ca
2、
原式=1*(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2²-1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^16-1)(2^16+1)(2^32+1)+1
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64