摘要:若n∈N.且n为奇数.则6n+Cn16n-1+-+Cnn-16-1被8除.所得的余数是 .5
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(2013•东莞一模)设等差数列{an},{bn}前n项和Sn,Tn满足
=
,且
+
=
,S2=6;函数g(x)=
(x-1),且cn=g(cn-1)(n∈N,n>1),c1=1.
(1)求A;
(2)求数列{an}及{cn}的通项公式;
(3)若dn=
,试求d1+d2+…+dn.
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| Sn |
| Tn |
| An+1 |
| 2n+7 |
| a3 |
| b4+b6 |
| a7 |
| b2+b8 |
| 2 |
| 5 |
| 1 |
| 2 |
(1)求A;
(2)求数列{an}及{cn}的通项公式;
(3)若dn=
|