数列{an}满足a(n+1)+(-1)^n*an=2n-1,则{an}的前60项和为? 大神
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/18 14:21:17
数列{an}满足a(n+1)+(-1)^n*an=2n-1,则{an}的前60项和为? 大神
数列{an}满足a(n+1)+(-1)^n*an=2n-1,则{an}的前60项和为? 大神帮忙啊!谢啦!
数列{an}满足a(n+1)+(-1)^n*an=2n-1,则{an}的前60项和为? 大神帮忙啊!谢啦!
当n=4K(k为正整数)时有:
a(4K)-a(4K-1)=a(4K)+(-1)^(4k-1)*a(4k-1)=2*(4k-1)-1=8k-3
a(4k-1)+a(4k-2)=a(4k-1)+(-1)^(4k-2)*a(4k-2)=2(4k-2)-1=8k-5
a(4k-2)-a(4k-3)=a(4k-2)+(-1)^(4k-3)*a(4k-3)=2(4k-3)-1=8k-7
那么:
a(4k-3)+a(4k-2)+a(4k-1)+a(4k)
=-[a(4k-2)-a(4k-3)]+2*[a(4k-1)+a(4k-2)]+[a(4K)-a(4K-1)]
=-(8k-7)+2*(8k-5)+(8k-3)
=16k-6
则:
S60=a1+a2+a3+...+a60
=[a1+a2+a3+a4]+[a5+a6+a7+a8]+.+[a57+a58+a59+a60]
=16*1-6+16*2-6+.+16*15-6
=16*(1+2+3+...+15)-15*6
=16*(1+15)*15/2-90
=1920-90
=1830
a(4K)-a(4K-1)=a(4K)+(-1)^(4k-1)*a(4k-1)=2*(4k-1)-1=8k-3
a(4k-1)+a(4k-2)=a(4k-1)+(-1)^(4k-2)*a(4k-2)=2(4k-2)-1=8k-5
a(4k-2)-a(4k-3)=a(4k-2)+(-1)^(4k-3)*a(4k-3)=2(4k-3)-1=8k-7
那么:
a(4k-3)+a(4k-2)+a(4k-1)+a(4k)
=-[a(4k-2)-a(4k-3)]+2*[a(4k-1)+a(4k-2)]+[a(4K)-a(4K-1)]
=-(8k-7)+2*(8k-5)+(8k-3)
=16k-6
则:
S60=a1+a2+a3+...+a60
=[a1+a2+a3+a4]+[a5+a6+a7+a8]+.+[a57+a58+a59+a60]
=16*1-6+16*2-6+.+16*15-6
=16*(1+2+3+...+15)-15*6
=16*(1+15)*15/2-90
=1920-90
=1830
数列{an}满足an+1=(-1)n*an+n,则{an}的前100项和为多少?
已知数列{an}满足a1=1,an-a(n+1)=ana(n+1),数列{an}的前n项和为Sn.(1)求证:{1/an
数列an中,an=1/(根号(n+2)+根号n),则an的前n项和为
已知数列{an}的前n项和为Sn,且满足Sn=2an-1,n为正整数,求数列{an}的通项公式an
已知数列{an}的前n项和为Sn,且满足Sn=2an-1(n属于正整数),求数列{an}的通项公式an
数列{An}满足A1=1,An+1=An/2An+1 数列Bn的前n项和为Sn=12-12(2/3)n
数列通式问题数列an的an=an-1+2^n(n>2 n∈N*)则它的通项公式数列an的前n项和Sn满足an=2-2Sn
数列{An}满足An+1 + (-1)的n次方乘an=2n—1.则an的前60项和
已知数列{an}满足a1=2,a(n+1)=(5an-13)/(3an-7)则数列{an}的前100项的和是
设数列{an}的前n项和为Sn,并且满足2Sn=an²+n,an>0.(1)求a1,a2,a3.(2)猜想{a
数列{an}满足a1=33,a(n+1)-an=2n,则an/n的最小值为_
已知数列An满足An>0,其前n项和为Sn为满足2Sn=An的平方+An(1)求An(2)设数列Bn满足An/2的n次方