n^2 n^3 (-2^n^2)等于?
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这道题用错位相减法.原式/2=1/4n+3/8n+...+(2n-1)/n*2^(n+1)所以原式/2=1/2n+2/4n+2/8n+...+2/n*2^n-(2n-1)/n*2^(n+1)n*原式/
lim[(n+3)/(n+1)]^(n-2)=lim[1+2/(n+1)]^(n-2)=lim{[1+2/(n+1)]^[(n+1)/2]}^[(n-2)×2/(n+1)]=lime^[2(n-2)/
n-1=再问:29/20n-4n=
这很简单就是整式的加减法和乘法,大约是初一(七年级)下学期的内容1+(n+1)+n*(n+1)+n*n+(n+1)+1=1+n+1+n²+n+n²+n+1+1=2n²+3
-n(n-4)^2
当n=k+1时,等式左边一共有k+1个式子相乘,倒数第二项就是[(k+1)+k],(k+k)这一项其实是(k+1+k-1)也就是倒数第三项.再问:那比如2+4+6+8+2nn=k+1左边=2+4+6+
根号(n^2+1)-nn+根号(n^2-1)所以根号(n^2+1)-n
原式=[n(n+3)[(n+1)(n+2)]+1=(n2+3n)[(n2+3n)+2]+1(n2+3n)2+2(n2+3n)+1=(n2+3n+1)2=n2+3n+1.
原式=(3n²+3n+2n²-3n²+n+6n²+12n)/6=(2n²+6n²+16n)/6=(n²+3n+8)/3
设n+2=x所以(n+1)(n+2)(n+3)=(x-1)*x*(x+1)=(x^2-1)*x=x^3-x将n+2=x代入,得n^3+3n^2*2+3n*2^2+2^3-n-2=n^3+6n^2+12
等式两侧同时除以2^n,所以要证的式子等价于:(3/2)^n-1>1也就是:(3/2)^n>2当n=2时,(3/2)^2=9/4>2,成立.而f(n)=(3/2)^n是个增函数,所以当n>2时,(3/
(2n+3n)的2次方-(6n+9n)原式=5n的2次方-15n=25n-15n=10n
un=(1/(n^2+n+1)+2/(n^2+n+2)+3/(n^2+n+3)……n/(n^2+n+n)),k/(n^2+n+n)≤k/(n^2+n+k)≤k/n^2==>(1+2+..+n)/(n^
证明:∵n∈N∴2^n=(1+1)^n=C(0,n)+C(1,n)+...+C(k,n)+...+C(n,n),(0<k<n,n,k∈N)∵n≥3∴2^n=C(0,n)+C(1,n)+...+C(n-
先证明对于任意x≠0,1+xf(0)=1>0,即1+x
1/2*f(1/2)=(1/2)^2+3*(1/2)^3...+(2n-1)*(1/2)^(n+1)f(1/2)-1/2*f(1/2)=1/2+2*(1/2)^2+2*(1/2)^3+...+2*(1
首先写出当n=k时和n=k+1时等式左边的式子,当n=k时,左边等于(k+1)(k+2)…(k+k)=(k+1)(k+2)…(2k),①当n=k+1时,左边等于(k+2)(k+3)…(k+k)(2k+
n/(n+2)!=(n+2-2)/(n+2)!=1/(n+1)!-2/(n+2)!.所以原式=1/1!-2/2!+1/2!-2/3!+.=1/1!-1/2!-1/3!-.=1-(e-2)=3-e.
可利用归纳法证明n=2时,2/1=2,成立假设n=2k时,k为正整数,结论成立则n=2k+2时,有(2k+2)/(2k+1)+(2k+2)(2k)/[(2k+1)(2k-1)]+...+(2k+2)(