1.求数列11,103,1005,10007.前n项和2.求数列1/1.5,1/3.7,1/5.9.的前n项和3.求数列（2^2+1）/（2^2-1）+（3^3+1）/（3^3-1)+（4^4+1）/（4^4-1)...的前n项和4.求1.3,2.5,3.7..的前n项和5.求1+（1+1/2）+(1+1/2+1/4)+...(1+1/2+1/4

Question

1.求数列11,103,1005,10007.前n项和
2.求数列1/1.5,1/3.7,1/5.9.的前n项和
3.求数列（2^2+1）/（2^2-1）+（3^3+1）/（3^3-1)+（4^4+1）/（4^4-1)...的前n项和
4.求1.3,2.5,3.7..的前n项和
5.求1+（1+1/2）+(1+1/2+1/4)+...(1+1/2+1/4+1/2^(n-1)).的前n项和
6.求2,4*2^2,7*2^3...的前n项和
每一道都要有过程,感激不尽.可以追分.

Answer 1

1.
11 = 10^1 + (2·1 - 1)
103 = 10^2 + (2·2 - 1)
1005 = 10^3 + (2·3 - 1)
10007 = 10^4 + (2·4 - 1)
∴an = 10^n + 2n - 1
10^1 + ...+ 10^n = 10(10^n - 1)/9
2·1 + 2·2 + ...+ 2·n = n^2 + n ,而（-1)·n = -n
∴Sn = [10(10^n - 1)/9] + n^2
2.
an = 1/[1.5 + 2.2(n-1)] = 10/(22n - 7)
无法求出前n项和 ,这和求(1 + 1/2 + 1/3 + 1/4 ...+ 1/n)道理相同
3.
该数列通项为：an = (n^n + 1)/(n^n - 1) = 1 + [2/(n^n - 1)]
而n^n - 1 = n^n - 1^n = (n-1){n^(n-1) + [n^(n-1)-1] + ...+ 1}
计算过于庞杂 ,很抱歉我只能算到这里
4.
3.7 - 2.5 = 2.5 - 1.3 = 1.2 ,∴这是以1.3为首项 ,1.2为公差的等差数列 ,∴an = 1.3 + 1.2(n-1) = 1.2n + 0.1 ,∴Sn = (a1 + an)·n/2 = (6n^2 + 7n)/10
5.
易得：通项an = [1+1/2+1/4+1/2^(n-1)]
= [(1/2)^0 + (1/2)^1 + (1/2)^2 + (1/2)^(n-1)]
= 2 - (1/2)^(n-1)
∴Sn = 2n - {[(1/2)^(1-1)] + [(1/2)^(2-1)] + ...+ [(1/2)^(n-1)]}
= 2n - 2 + (1/2)^(n-1)
6.
a1 = 1·2^1 ,a2 = (1+3)·2^2 ,a3 = (1 + 3·2)·2^3
∴通项an = (3n - 2)·2^n = 3n·2^n - 2^(n+1)
∴Sn =[3·2 + 6·4 + 9·8 + ...+ 3n·2^n]-[2^2 + ...+ 2^(n+1)]
令Tn = [3·2 + 6·4 + 9·8 + ...+ 3n·2^n]
令Kn = [2^1 + 2^2 + ...+ 2^(n+1)]
易求得：Kn = 2^(n+2) - 4
现用拆项相消法求 Tn ：
2Tn = 3·4 + 6·8 + 9·16 + ...+ 3n·2^(n+1)
∴Tn = 2Tn - Tn = [3n·2^(n+1) - 3·2] - 3·(4 + 8 + 16 + ...+ 2^n)
= [3n·2^(n+1) - 3·2] - 3·[2^(n+2) - 4]
= 3(n-1)·2^(n+1) + 6
∴Sn = Tn - Kn
= (3n - 5)·2^(n+1) + 10