试求(2+1)(2²+1)(2的四次方+1）...（2的2n次方+1）+1的值.

(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^2n+1)=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)...(2^2n+1) 　　　原式乘以1,即乘以(2-1),值不变=(2^2-1)(2^2+1)(2^4+1)...(2^2n+1) 往下递次使用平方差公式(a-1)(a+1)=a^2-1=(2^4-1)(2^4+...