∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)
来源：学生作业帮助网 编辑：六六作业网 时间：2024/07/03 09:03:33
∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)a1=S1=1-2+
∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)
∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)
∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)
a1=S1=1-2+3=2
Sn=n²-2n+3
Sn-1=(n-1)²-2(n-1)+3
an=Sn-Sn-1=n²-2n+3-(n-1)²+2(n-1)-3=2n-3
n=1时,2-3=-1,与a1=2矛盾,因此n=1时,a1=2
n≥2时,an=2n-3
数列通项公式为
an=2 n=1
2n-3 n≥2
∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6)
∞∑ n=1 [(2n-1)/(3^n)]*[x^(2n-2)] ∞∑ n=1 [((-1)^n)/(2n-1)*(x^(2n)
证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)...1=n
判断敛散性：∑(n=1~∞) (4^n*n!*n!)/(2n)!
∑（n^2-n^3/2^n+3^n)求证他是绝对收敛 n=1
设f(n)=1+2+3+.n,则(n-->+∞)limf(n)/[f(n)]=
高数题：n趋近于0,lim｛1/（n^2+n+1)+2/(n^2+n+2)+3/(n^2+n+3)+.+n/(n^2+n+n)｝=?
为什么n(n+1)(n+2)(n+3)=(n²+3n+2)(n²+3n)?
为什么:n×(n+1)=1/3[n(n+1)×(n+2)-(n-1)×n×(n+1)]
判定级数∞∑n=1 [(-1)^n-1]*(3^n)(x^2n)/n]的敛散性.
lim (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中n→∞
数列a(n)=n (n+1)(n+2)(n+3),求S(n)
n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10这道题怎么解
n(n-1)(n-2)（n-3)（n-4)=154440.求N值 要步骤
想向您请教 ∞∑ n=0（-1）^n*(n^2-n+1)/2^n求和及∞∑ n=1 2^n/(3^n(2n-1))求和
1*N+2*（N-1）+3*（N-2）+...+N*1=1/6N(N+1)(N+2)
用数学归纳法证明：(n+1)(n+2)(n+3)+.+(n+n)=(2^n)*1*3*.(2n-1)