已知正项数列{an}满足a1=32,a2n+1−a2n=1(n+2)2−1n2,记数列{an}的前n项和为Sn,则1S1−1S3+1S5−⋯−1S2007+1S2009的值为______.
分析与解 10062011.
由累加法知a2n=1(n+1)2+1n2+1,
所以an=1+1n−1n+1.直接求和得Sn=n+1−1n+1,所以1Sn=n+1n(n+2)=12(1n+1n+2).
所求的式子记为M,有2M=1+13−(13+15)+(15+17)−(17+19)+⋯+(12009+12011)=20122011,
所以M=10062011.