在数列中,已知a1=1,an+1-an=sin(n+1)π/2,记Sn为数列an的前n项和,则S2014=网上有这样的答案a1=1,a(n+1)-an=sin(n+1)π/2a2-a1=-1,a2=0 a3-a2=0,a3=a2=0a4-a3=1,a4=1 a5-a4=0,a5=a4=1得数列1,0,0,1,1,0,0,1,1,0..S2014=2012/2+1=1007可
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![在数列中,已知a1=1,an+1-an=sin(n+1)π/2,记Sn为数列an的前n项和,则S2014=网上有这样的答案a1=1,a(n+1)-an=sin(n+1)π/2a2-a1=-1,a2=0 a3-a2=0,a3=a2=0a4-a3=1,a4=1 a5-a4=0,a5=a4=1得数列1,0,0,1,1,0,0,1,1,0..S2014=2012/2+1=1007可](/uploads/image/z/3622240-64-0.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5a1%3D1%2Can%2B1-an%3Dsin%28n%2B1%29%CF%80%2F2%2C%E8%AE%B0Sn%E4%B8%BA%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E5%88%99S2014%3D%E7%BD%91%E4%B8%8A%E6%9C%89%E8%BF%99%E6%A0%B7%E7%9A%84%E7%AD%94%E6%A1%88a1%3D1%2Ca%EF%BC%88n%2B1%EF%BC%89-an%3Dsin%28n%2B1%29%CF%80%2F2a2-a1%3D-1%2Ca2%3D0+a3-a2%3D0%2Ca3%3Da2%3D0a4-a3%3D1%2Ca4%3D1+a5-a4%3D0%2Ca5%3Da4%3D1%E5%BE%97%E6%95%B0%E5%88%971%2C0%2C0%2C1%2C1%2C0%2C0%2C1%2C1%2C0..S2014%3D2012%2F2%2B1%3D1007%E5%8F%AF)
在数列中,已知a1=1,an+1-an=sin(n+1)π/2,记Sn为数列an的前n项和,则S2014=网上有这样的答案a1=1,a(n+1)-an=sin(n+1)π/2a2-a1=-1,a2=0 a3-a2=0,a3=a2=0a4-a3=1,a4=1 a5-a4=0,a5=a4=1得数列1,0,0,1,1,0,0,1,1,0..S2014=2012/2+1=1007可
在数列中,已知a1=1,an+1-an=sin(n+1)π/2,记Sn为数列an的前n项和,则S2014=
网上有这样的答案
a1=1,
a(n+1)-an=sin(n+1)π/2
a2-a1=-1,a2=0 a3-a2=0,a3=a2=0
a4-a3=1,a4=1 a5-a4=0,a5=a4=1
得数列1,0,0,1,1,0,0,1,1,0..S2014=2012/2+1=1007
可是a2-a1=sin2π/2=0 不等于-1啊
最后得的数列应该是1,1,0,0,1,1,0,0...
在数列中,已知a1=1,an+1-an=sin(n+1)π/2,记Sn为数列an的前n项和,则S2014=网上有这样的答案a1=1,a(n+1)-an=sin(n+1)π/2a2-a1=-1,a2=0 a3-a2=0,a3=a2=0a4-a3=1,a4=1 a5-a4=0,a5=a4=1得数列1,0,0,1,1,0,0,1,1,0..S2014=2012/2+1=1007可
网上有错误,但原理可取.
为防止混淆,下标一律加括号.
a(1)=1
a(2)-a(1)=sinπ=0 a(2)=a(1)=1
a(3)-a(2)=sin(3π/2)=-1 a(3)=a(2)-1=0
a(4)-a(3)=sin(4π/2)=0 a(4)=a(3)+0=0
a(5)-a(4)=sin(5π/2)=1 a(5)=a(4)+1=1
……
a(n+1)-a(n)=sin(n+1)π/2 sin(n+1)π/2从n=1开始,以0,-1,0,1的次序循环,
所以最后得的数列应该是1,1,0,0,1,1,0,0...,每连续4个(为组)的和为2, 2014/4=503+2/4,前2014个数分503组,和为1006,剩余两个为一组的前两个,和为2,所以总和 为1008.