【答案】
分析:(1)由題意數(shù)列{a
n} 的首項(xiàng)為a
1=1,前n項(xiàng)和為S
n,且na
n-S
n=2n(n-1),利用數(shù)列的前n項(xiàng)和求出通項(xiàng)即可;
(2)①有數(shù)列 {b
n} 滿足:4b
n=S
n+n-1+(-1)
n,先推導(dǎo)出

通項(xiàng)公式,②并對(duì)該式子分奇偶進(jìn)行討論求出2n-

,并有導(dǎo)出

的通項(xiàng)公式代入,再利用數(shù)列的極限求得.
解答:解:(1)因?yàn)橛幸阎簄a
n-S
n=2n(n-1),a
2=5,
當(dāng)n≥2時(shí),(n-1)a
n-1-S
n-1=2(n-1)(n-2),
∴na
n-(n-1)a
n-1-S
n+S
n-1=2n(n-1)-2(n-1)(n-2),
即(n-1)(a
n-a
n-1)=4(n-1)(n≥2),∴a
n-a
n-1=4(n≥2),
故數(shù)列{a
n}是公差為4的等差數(shù)列,
∴a
n=4n-3(n∈N
+);
(2)由于數(shù)列 {b
n} 滿足:4b
n=S
n+n-1+(-1)
n,
∴4b
n=2n
2-1+(-1)
n(n∈N
+),∴

,
故

,
當(dāng)n為大于0的偶數(shù)時(shí),

,
當(dāng)n為大于1的奇數(shù)時(shí),

,
∴E
9=(b
1+b
3+b
5+b
7+b
9)+(b
2+b
4+b
6+b
8)=

當(dāng)n>1,且n∈N
+時(shí),若n為偶數(shù),則

,
若n為大于1的奇數(shù),則

,
∴
∴

.
點(diǎn)評(píng):此題考查了學(xué)生的分類討論的能力及嚴(yán)謹(jǐn)?shù)倪壿嬐茖?dǎo)能力,還考查了已知數(shù)列的前n項(xiàng)的和求數(shù)列的通項(xiàng),數(shù)列的極限.