3^1=3,3^=9,3^3=27.3^8=6561
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/16 12:18:07
3^1=3,3^=9,3^3=27.3^8=6561
反复运用平方差公式:
(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^8-1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^16-1)(3^16+1)(3^32+1)+2
=(3^32-1)(3^32+1)+2
=(3^64-1)+2
=3^64+1
注意到3的幂的个位数是按照3、9、7、1依次循环的,故3^64的个位数与3^4的个位数相同是1
因此所求个位数是3
再问: 答案不对啊
再答: 写错了,所求个位数是1+1=2
(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^8-1)(3^8+1)(3^16+1)(3^32+1)+2
=(3^16-1)(3^16+1)(3^32+1)+2
=(3^32-1)(3^32+1)+2
=(3^64-1)+2
=3^64+1
注意到3的幂的个位数是按照3、9、7、1依次循环的,故3^64的个位数与3^4的个位数相同是1
因此所求个位数是3
再问: 答案不对啊
再答: 写错了,所求个位数是1+1=2
3^1=3,3^=9,3^3=27.3^8=6561
1+3+9+27+81+.+6561=________
观察下列算式:31=3,32=9,33=27,34=81,35=243,36=729,37=2187,38=6561;那
已知31=3,32=9,33=27,34=81,35=243,36=729,37=2187,38=6561…
观察下列算式:31=3,32=9,33=27,34=81,35=243,36=729,37=2187,38=6561,…
3的1次方是3,2次方是9,3的3次方=27.3的2015的末尾数是多少,是怎么解出的,
求出未知数的值.4x-3*2.7=27.3
x-1/3x=8/9
1+2+3+4、、、+8+9=
2 3 6 9 17 () 1 3 3 8 ? 8=(3-1)(3+1) ?=(8-1)(3+1) =28 2 3 6
6+1=4+[ ]=[ ]+[ ]=[ ] 8-3=6-[ ]=7-[ ]=9-[ ] 5+4=3+[ ]=[ ]+[
已知3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=656