1.求Sn=2/2+3/2∧2+4/2∧3+...+n/2∧n-1+n+1/2∧n

1.Sn/2=2/2^2+3/2^3+...+(n+1)/2^(n+1)
=3/2∧2+4/2∧3+...+n/2∧n-1+n+2/2∧(n+1)-[1/2^2+1/2^3+1/2^4+...+1/2^(n+1)]
=S(n+1)-2/2-[1/2-1/2^(n+1)]
=Sn+(n+2)/2^(n+1)-3/2+1/2^(n+1)
=Sn+(n+3)/2^(n+1)-3/2
因此Sn/2=3/2-(n+3)/2^(n+1)
Sn=3-(n+3)/2^n
2.前n项包括的奇数的个数为：1+2+3+...+n=n(n+1)/2
因为前n个奇数之和是n^2,所以数列1,3+5,7+9+11,13+15+17+19,.的前N项和为：
[n(n+1)/2]^2=n^2(n+1)^2/4