Question

求大神解决两道数学题,非常感谢.

Answer 1

（1）构造平方差公式

原式=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1

={[(2+1)(1-2)][(2^2+1)(1-2^2)][(2^4+1)(1-2^4)][(2^8+1)(1-2^8)][(2^16+1)(1-2^16)]/[(1-2)(1-2^2)(1-2^4)(1-2^8)(1-2^16)]}+1

=(1-2^2)(1-2^4)(1-2^8)(1-2^16)(1-2^32)/[(1-2)(1-2^2)(1-2^4)(1-2^8)(1-2^16)]+1

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

=2^32

（2）将x=1代入得(2k+a)/3=2+(1-bk)/6

化简得(4+b)k=13-2a,因为与k无关,故应消去k,即4+b=0且13-2a=0,故b=-4,a=13/2.