数列an的前n项和为sn,存在常数A,B,C使得an+sn=An^2+Bn+C对任意正整数n都成立.1,数列an是等差数列,(1)若an=2n-1,求A,B,C的值.(2)若C=0,a1=1,bn=1/an*an+1,P=∑(1+bn) (∑上面是2013,下面是i=1)2,若A=-1/2,B=
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![数列an的前n项和为sn,存在常数A,B,C使得an+sn=An^2+Bn+C对任意正整数n都成立.1,数列an是等差数列,(1)若an=2n-1,求A,B,C的值.(2)若C=0,a1=1,bn=1/an*an+1,P=∑(1+bn) (∑上面是2013,下面是i=1)2,若A=-1/2,B=](/uploads/image/z/2771733-21-3.jpg?t=%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E5%AD%98%E5%9C%A8%E5%B8%B8%E6%95%B0A%2CB%2CC%E4%BD%BF%E5%BE%97an%2Bsn%3DAn%5E2%2BBn%2BC%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0n%E9%83%BD%E6%88%90%E7%AB%8B.1%2C%E6%95%B0%E5%88%97an%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%EF%BC%881%EF%BC%89%E8%8B%A5an%3D2n-1%2C%E6%B1%82A%2CB%2CC%E7%9A%84%E5%80%BC.%EF%BC%882%EF%BC%89%E8%8B%A5C%3D0%2Ca1%3D1%2Cbn%3D1%2Fan%2Aan%2B1%2CP%3D%E2%88%91%EF%BC%881%2Bbn%EF%BC%89+%EF%BC%88%E2%88%91%E4%B8%8A%E9%9D%A2%E6%98%AF2013%2C%E4%B8%8B%E9%9D%A2%E6%98%AFi%3D1%EF%BC%892%2C%E8%8B%A5A%3D-1%2F2%2CB%3D)
数列an的前n项和为sn,存在常数A,B,C使得an+sn=An^2+Bn+C对任意正整数n都成立.1,数列an是等差数列,(1)若an=2n-1,求A,B,C的值.(2)若C=0,a1=1,bn=1/an*an+1,P=∑(1+bn) (∑上面是2013,下面是i=1)2,若A=-1/2,B=
数列an的前n项和为sn,存在常数A,B,C使得an+sn=An^2+Bn+C对任意正整数n都成立.
1,数列an是等差数列,
(1)若an=2n-1,求A,B,C的值.
(2)若C=0,a1=1,bn=1/an*an+1,P=∑(1+bn) (∑上面是2013,下面是i=1)
2,若A=-1/2,B=-3/2,C=1,设Cn=an+n
(1)求证:数列cn为等比数列.
(2)求数列nCn的前n项和Tn.
数列an的前n项和为sn,存在常数A,B,C使得an+sn=An^2+Bn+C对任意正整数n都成立.1,数列an是等差数列,(1)若an=2n-1,求A,B,C的值.(2)若C=0,a1=1,bn=1/an*an+1,P=∑(1+bn) (∑上面是2013,下面是i=1)2,若A=-1/2,B=
1.(1).若an=2n-1,则Sn=n^2,所以2n-1+n^2=An^2+Bn+C,对比系数,A=1,B=2,C=-1;
(2)若C=0,a1=1,设an=1+(n-1)d=nd-d+1.所以Sn=n+n(n-1)d/2,所以,Sn+an=(d/2+1)n+d/2n^2-d+1,因为C=0,所以d=1,所以an=n,所以bn=1/an*(an+1),所以P=(1+1+1+.+1)+(1-1/2+1/2-1/3+.-1/2014)=2013+2013/2014=4096195/2014;
2.(1)因为an+Sn=An^2+Bn+C .1
所以a(n-1)+S(n-1)=A(n-1)^2+B(n-1)+C.2 『小括号里是下标』
1式减2式,得
2an=a(n-1)-n-1
两边加上2n,得2(an+n)=a(n-1)+n-1
即2Cn=C(n-1)
所以是公比为1/2的等比数列;
(2)求a1
a1=S1
所以2a1=-1,a1=-1/2
所以an=-1/2*(1/2)^(n-1)
所以Tn=-1/2*(1+2*1/2+3*(1/2)^2+4*(1/2)^3+.+n*(1/2)^(n-1))
1/2Tn=-1/2*(1*1/2+2*(1/2)^2+3*(1/2)^3+4*(1/2)^4+.+n*(1/2)^n)
上式减下式,得
1/2Tn=-1/2*(1+1/2+(1/2)^2+.+(1/2)^(n-1)-n*(1/2)^n)=-1/2*(2-(1/2)^(n-1)-n*(1/2)^n)
乘上2,得Tn=2-(2+n)*(1/2)^n OK