已知{an}為等差數(shù)列,{bn}為等比數(shù)列,a1=b1=1,a5=5(a4-a3),b5=4(b4-b3).
①求{an}和{bn}的通項(xiàng)公式;
②設(shè)cn=an+bn,求數(shù)列的前n項(xiàng)和Rn.
③設(shè)tn=anbn,求數(shù)列{tn}的前n項(xiàng)和Tn.
④記{an}的前n項(xiàng)和為Sn,求證:SnSn+2<S2n+1(n∈N*).
S
2
n
+
1
【考點(diǎn)】錯位相減法.
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發(fā)布:2024/9/6 11:0:13組卷:89引用:1難度:0.5
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