(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)…(2^2n+1)
2个回答
原式乘以(2-1)
然后一直用平方差公式
原式等于(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)…(2^2n+1)
=2^4n-1
相关问题
{2+1/2} {2^2+1/2^2} {2^4+1/2^4}{2^8+1/2^8}{2^16+1/2^16}
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)-(4^2)^8-(-1)^2011
求和:1/2-1/4+1/8-1/16+……+(-1)^n-1*1/2^n
前n项求和求和Sn=1(1\2)+2(1\4)+3(1\8)+4(1\16)+…+n(1\2^n)这里的1(1\2)、2
(1+2)(1+2^2/1)(1+2^4/1)(1+2^8/1)+2^16/1
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^16
计算(1+2)(1+2∧2)(1+2∧4)(1+2∧8)(1+2∧16)=?
(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+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)-2^32