an的通项公式an=(-1)n-1次方
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an+1=an+2n推出an=an-1+2(n-1)...a2=a1+2累加得an=a1+2(2+3+4+...n-1)an=2+n(n-1)an=n^2-n+2(n>=1)
lnan=ln2根号(a(n-1))lnan=ln2+(1/2)lna(n-1)2lnan=ln4+lna(a-1)2(lnan-ln4)=lna(n-1)-ln4令bn=lnan-ln4所以{bn}
∵an=1n(n+1)=1n−1n+1∴Sn=a1+a2+…+an=1−12+12−13+…+1n−1n+1=1−1n+1=nn+1∴nn+1=1011∴n=10故答案为:10
n>=2S(n-1)=n/(n-1)所以an=Sn-S(n-1)=-1/(n²-n)a1=S1=2/1=2所以an=2,n=1-1/(n²-n),n≥2
A1=2A(n+1)-An=nAn=[An-A(n-1)]+[A(n-1)-A(n-2)]+…+(A2-A1)+A1=(n-1)+(n-2)+…+2+1+2=(n-1)*n/2+2=(n^2-n+4)
1/an-an=2√n且an>0,(an)^2+2√n(an)-1=0,(an)=[-2√n+√(4n+4)]/2=-√n+√(n+1).而,(an)=[-2√n-√(4n+4)]/2=-√n-√(n
an=Sn-Sn-1=n(n-1)-(n-1)(n-2)=2n,而a1=2×1=S1=1×(1+1)=2,即n=1时也符合条件;故an=2n
此类题目采用累加法或迭代法∵an+1-an=3n(往下递推)∴an-an-1=3(n-1)an-1-an-2=3(n-2).a3-a2=3×2a2-a1=3×1以上格式左边+左边=右边+右边左边相加的
由题意得an^2+2根号n*an-1=0解出来以后讨论下,因为an>0an=-根号下n+根号下n+1
a(n+1)=-an+3n-54a(n+1)+x(n+1)+y=-an+3n-54+x(n+1)+ya(n+1)+x(n+1)+y=-[an-(3+x)n+54-x-y]令x=-(3+x)y=54-x
由an+1-an=3n,可知a2−a1=3a3−a2 =6…an−an−1=3(n−1)将上面各等式相加,得an-a1=3+6+…+3(n-1)=3n(n−1)2∴an=a1+3n(n−1)
首先,我敢说你一定除反了[(3n+3)an+4n+6》/n才对(我做过嘻嘻)把n乘过去,左右加2n得n(an+1+2)=(n+1)3(an+2)很似曾相识吧同除n(n+1)(an+1+2)/(n+1)
将已知等式取倒数,得1/an=[3a(n-1)+1]/a(n-1)=1/a(n-1)+3,所以,{1/an}是首项为1/a1=1,公差为3的等差数列,因此1/an=1+3(n-1)=3n-2,所以an
An=1/n(n+1)=1/n-1(n+1)S5=a1+a2+a3+a4+a5=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6=1-1/6=5/6A
A_{n}+A_{n+1}-1=n*(A_{n+1}-A_{n-1})-------------------------1A_{n-1}+A_{n}-1=(n-1)*(A_{n}-A_{n-2})--
题目怎么理解,是(3n-1)/(2n+1)再问:对的再答:哦这个就好做了的|(3n-1)/(2n+1)-3/2|=|-5/{2[2n+1]}|=2.5/[2n+1]
a(n+1)=2an/(an+2)1/a(n+1)=(an+2)/(2an)=1/an+1/21/a(n+1)-1/an=1/2,为定值.1/a1=1/1=1数列{1/an}是以1为首项,1/2为公差
an-a(n-1)=2na(n-1)-a(n-2)=2(n-1)a(n-2)-a(n-3)=2(n-2).a2-a1=2X2=4把以上n-1项相加得:an-a1=n^2+n-2解得:an=n^2+n