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(Ⅰ)∵Sn=2an+n2-3n-2,
∴Sn+1=2an+1+(n+1)2-3(n+1)-2.
∴an+1=2an-2n+2,∴an+1-2(n+1)=2(an-2n).
∴{an-2n}是以2为公比的等比数列;
(Ⅱ)a1=S1=2a1-4,∴a1=4,∴a1-2×1=4-2=2.
∴an-2n=2n,∴an=2n+2n.
当n为偶数时,Pn=b1+b2+b3+…+bn
=(b1+b3+…+bn-1)+(b2+b4+…+bn)
=-(2+2×1)-(23+2×3)-…-[2n-1+2(n-1)]+(22+2×2)+(24+2×4)+…+(2n+2×n)
=
?
+n=
?(2n?1)+n;
当n为奇数时,Pn=?
?(n+1).
综上,Pn=
;
(Ⅲ)cn=
=
.
当n=1时,T1=
<
当n≥2时,Tn=
+
+
+…+
<
+
+
+…+
=
+
=
+
?
=
?
<
<
综上可知:任意n∈N,Tn<
.
∴Sn+1=2an+1+(n+1)2-3(n+1)-2.
∴an+1=2an-2n+2,∴an+1-2(n+1)=2(an-2n).
∴{an-2n}是以2为公比的等比数列;
(Ⅱ)a1=S1=2a1-4,∴a1=4,∴a1-2×1=4-2=2.
∴an-2n=2n,∴an=2n+2n.
当n为偶数时,Pn=b1+b2+b3+…+bn
=(b1+b3+…+bn-1)+(b2+b4+…+bn)
=-(2+2×1)-(23+2×3)-…-[2n-1+2(n-1)]+(22+2×2)+(24+2×4)+…+(2n+2×n)
=
4(1?2n) |
1?22 |
2(1?2n) |
1?22 |
2 |
3 |
当n为奇数时,Pn=?
2n+1+2 |
3 |
综上,Pn=
|
(Ⅲ)cn=
1 |
an?n |
1 |
2n+n |
当n=1时,T1=
1 |
3 |
37 |
44 |
当n≥2时,Tn=
1 |
21+1 |
1 |
22+2 |
1 |
23+3 |
1 |
2n+n |
<
1 |
3 |
1 |
22 |
1 |
23 |
1 |
2n |
=
1 |
3 |
| ||||
1?
|
1 |
3 |
1 |
2 |
1 |
2n |
5 |
6 |
1 |
2n |
5 |
6 |
37 |
44 |
综上可知:任意n∈N,Tn<
37 |
44 |
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