计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

1个回答

  • 思路:在原式乘上(2-1),不断的产生平方差,可以巧解.

    (2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

    原式=(2-1)(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

    =(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

    =(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

    =(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)

    =(2^16-1)(2^16+1)(2^32+1)(2^64+1)

    =(2^32-1)(2^32+1)(2^64+1)

    =(2^64-1)(2^64+1)

    =2^128-1