1.1/8,1/6,9/22,27/40,( ) A.27/16 B.27/14 C.81/40 D.81/44 求详解
问题描述：
1.1/8,1/6,9/22,27/40,( ) A.27/16 B.27/14 C.81/40 D.81/44 求详解
1.1/8,1/6,9/22,27/40,( )
A.27/16 B.27/14 C.81/40 D.81/44
2.3,4,12,18,44,( )
A.44 B.56 C.78 D.79
3.4,5,15,6,7,35,8,9,( )
A.27 B.15 C.72 D.63
4.1526,4769,2154,5397
A.2317 B.1545 C.1469 D.5213
答
1.原式通分为：1/8,3/18,9/22,27/40,由此可得最后一项分子为81.分子规律为：3^2-1,4^2+2,5^2-3,6^2+4,最后项分母为7^2-5=44.所以选D.
2.此时规律为 3*2-2=4,4*2+4=12,12*2-6=18,18*2+8=44,最后项为44*2-10=78.选C.
3.15=(4-1)*5,35=(5-1)*7,最后项为 (8-1)*9=63.所以选D.
4.该式各项的四位数的个位+千位=十位+百位.1+6=5+2,4+9=7+6,2+4=5+1,5+7=3+9,所以只有C符合这种关系,选C.