-1.你先去括号,就会发现规律,即 2的63次方加到2的0次方,再减去2的64次方.最后用一下等比数列求和公式.
(2+1)(2∧2+1)(2∧4+1)(2∧8+1)(2∧16+1)(2∧32+1)-2∧64求值
2个回答
相关问题
-
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
-
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
-
求值:(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)-2^32
-
计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
-
求解以下两式的值：1、（2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64
-
(1)已知m=(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)*(2^64
-
1/2+1/4+1/8+1/16+1/32+1/64简便运算＝﹙1／2＋1／4＋1／8＋1／16＋1／32＋1／64＋1
-
已知,A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^12
-
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字.
-
已知；(2^-1)^x=1/2*4^-1*1/8*16^-1*1/32*64^-1,试求x的值