运用平方差公式计算(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)^是乘方

问题描述:

运用平方差公式计算(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)
^是乘方

(3+1)(3^2+1)(3^4+1)(3^8+1)...[3^(2^n)+1]
=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)...[3^(2^n)+1]/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...[3^(2^n)+1]/2
=(3^4-1)(3^4+1)(3^8+1)...[3^(2^n)+1]/2
=……
={[3^(2^n)]^2-1}/2
={3^[2^(n+1)]-1}/2

(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)
=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^2n+1)/2
=(3^4-1)(3^4+1)(3^8+1)...(3^2n+1)/2
……
=(3^4n-1)/2