已知通项公式an=n的平方减n减2,写出前五项
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![已知通项公式an=n的平方减n减2,写出前五项](/uploads/image/f/4282718-14-8.jpg?t=%E5%B7%B2%E7%9F%A5%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3Dn%E7%9A%84%E5%B9%B3%E6%96%B9%E5%87%8Fn%E5%87%8F2%2C%E5%86%99%E5%87%BA%E5%89%8D%E4%BA%94%E9%A1%B9)
Sn=2n²+3n+1①则Sn-1=2(n-1)²+3(n-1)+1②(n>1)①式减②式得an=4n+1(n>1)当n=1a1=s1=2+3+1=6不符合an的式子故an=6(n
Sn=n^2+nS(n-1)=(n-1)^2+n-1=n^2-nan=Sn-S(n-1)=2nbn=1/2^an+n=1/2^(2n)+n=4^(-n)+n
n=1时,S1=a1=1+1=2n≥2时,Sn=n^2+1S(n-1)=(n-1)^2+1an=Sn-S(n-1)=n^2+1-(n-1)^2-1=2n-1n=1时,a1=2-1=1,与a1=2矛盾.
a1=S1=1-48=-47n>=2:an=Sn-S(n-1)=[n^2-48n]-[(n-1)^2-48(n-1)]=n^2-48n-(n^2-2n+1-48n+48)=2n-49a1=2*1-49
Sn=3n²+n+1①n=1时S1=a1=3+1+1=5n>=2时S(n-1)=3(n-1)²+(n-1)+1②①-②Sn-S(n-1)=an=3n²+n+1-[3(n-
当n=1时a1=s1=1²+1=2当n≥2时sn=n²+n---------------------------①s(n-1)=(n-1)²+(n-1)---------
上面答案要补充一下…求出an=2n-1时要验证当S1=a1=1这样才完整,如果S1不等于a1那么an就要就要以分段函数的形式来写咯
1:Sn=2n^2-3n+1Sn-1=2n^2-7n+6an=Sn-Sn-1=4n-52:a1+a1q^3=18a1q+a1q^2=12(1+q^3)/(q+q^2)=3/2(q-2)(2q^2+q-
因为Sn=3n²+n-1可得,S(n-1)=3(n-1)²+n-1-1=3n²-5n+1(n>2,且n∈N+)因此,an=Sn-S(n-1)=3n²-(3n^2
Sn=-3n²+22n+1an=Sn-S(n-1)=(-3n²+22n+1)-[-3(n-1)²+22(n-1)+1]=-3n²+22n+1+3n²-
Sn=n平方+2nS(n-1)=(n-1)²+2(n-1)an=Sn-S(n-1)=[n²-(n-1)²]+[2n-2(n-1)]=(n+n-1)(n-n+1)+2(n-
an=Sn-S[n-1]=2n^2-n+1-2(n^2-2n+1)+n-1-1=4n-3,(当n>=2)a1=S1=2经检验a1不符合通项an=4n-3所以an通项公式为2(当n=1)an=4n-3(
an=Sn-S(n-1)=-2n^2-n-[-2(n-1)^2-(n-1)]=2n^2-4n+2+n-1-(2n^2+n)=-4n+1
(1)an=Sn-Sn-1=2n-2(2)bn=2^[2(n-1)]+1=4^(n-1)+1令Cn=4^(n-1),Un={Cn}前n项的和.显然{Cn}是等比数列,∴Un=(4^n-1)/(4-1)
an=1-1/(n+1)
令b[n]=a[2n],c[n]=a[2n+1]b[n],c[n]均是等差数列直接用求和公式再反带回去
再问:你有一个地方学错了再问:是n大于等于2再问:已知数列an的前n项和Sn等于三分之二an减3,求an的通项公式
答:等差数列An=1+2nBn=(An)^2-1=(An-1)(An+1)=2n(2n+2)=4n(n+1)=4n^2+4nSn=4*[(1^2+2^2+3^2+...n^2)+(1+2+3+...+
因为Sn=n^2+1a1=s1=2∴S(n-1)=(n-1)^2+1∴an=Sn-S(n-1)=n^2+1-(n-1)^2-1=2n-1n≥2,且n∈N*∴an=2n=12n-1n≥2,且n∈N*
an=(3n-2)/(3n+1)a10=(3*10-2)/(3*10+1)=28/31(3n-2)/(3n+1)=7/107(3n+1)=10(3n-2)21n+7=30n-2030n-21n=7+2