(2+1)(2^2+1)(2^4+1)...(2^n+1)得多少?
来源:百度知道 编辑:UC知道 时间:2024/06/28 07:59:27
=(2-1)(2+1)(2^2+1)(2^4+1)……(2^n+1)/(2-1)
=(2^2-1)(2^2+1)(2^4+1)……(2^n+1)/1
=(2^4-1)(2^4+1)……(2^n+1)
=(2^8-1)(2^8+1)……(2^n+1)
=2^2n-1
=(2-1)(2+1)(2^2+1)(2^4+1)...(2^n+1)
=(2^2-1))(2^2+1)(2^4+1)...(2^n+1)
=...
=2^(2n)-1
我估计你想问的是和吧,是和的话是一个等比数列加一个等差数列,很好写,是乘积的话也可以写,太麻烦了,可能要用到二项式定理展开,
(1-√2)^2+(√2-1)^2(√2-1)^2+(-√2-1)^2
1( )2( )
1^2+2^2+...+n^2=?
(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)(2^10+1)......(2^2004+1)
2+2^1+2^2+2^3+...+2^2006=?
(2+1)(2*2+1)(2*2*2*2+1)......(2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2+1)的解法
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^32+1)
(2+1)(2^+1)(2^+1)(2^+1)(2^+1)(2^+1)(2^+1)的值为?
1+1大于2
计算(2+1)(2^2+1)(2^4+1)+。。。。+(2^2n+1)