求解这道数列题
1个回答
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an=a1+(n-1)d
9a5=6/5
9a1+36d =6/5
a1+4d =2/15 (1)
S6=4/5
3(2a1+5d)=4/5
2a1+5d = 4/15 (2)
2(1)-(2)
3d=0
d=0
from (1)
a1+ 0 = 2/15
a1 = 2/15
ie
an =2/15
bn
= S15 .n!/(n+3)!
=15(2/15).n!/(n+3)!
=2.n!/(n+3)!
=2/[(n+1)(n+2)(n+3)]
= 1/[(n+1)(n+2)] -1/[(n+2)(n+3)]
Tn
=b1+b2+..+bn
= 1/6 -1/[(n+2)(n+3)]
<1/6
9a5=6/5
9a1+36d =6/5
a1+4d =2/15 (1)
S6=4/5
3(2a1+5d)=4/5
2a1+5d = 4/15 (2)
2(1)-(2)
3d=0
d=0
from (1)
a1+ 0 = 2/15
a1 = 2/15
ie
an =2/15
bn
= S15 .n!/(n+3)!
=15(2/15).n!/(n+3)!
=2.n!/(n+3)!
=2/[(n+1)(n+2)(n+3)]
= 1/[(n+1)(n+2)] -1/[(n+2)(n+3)]
Tn
=b1+b2+..+bn
= 1/6 -1/[(n+2)(n+3)]
<1/6
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