1 2*3^n-1 2*3^n-1

用定积分来做把分母上提出个n^2,所以原极限=lim1/n*∑1/[(1+3(k/n)^2]=∫[1/(1+3x^2)]dx积分区间o到1=1/√3arctan√3x|(o到1)=1/√3（π/3-0

1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)=1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)=1/(n+1)-1/(n+

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)/

(2^n-1)/(2^n+1)>n/(n十1)(n≥3,n∈N+),1-2/(2^n+1)>1-1/(n+1),2/(2^n+1)

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

二项式展开,左=1+n*2/n+n(n+1)/2*(2n)²+.>=3+2(n+1)/n=5+2/n>5-2/nn>=3用在左边展开时,至少得到三项的合理性

其实就是1~12的平方和减去1~10的平方和n1=12,代入公式得到结果1n2=10,代入公式得到结果2减一下,就是最终结果了.

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

∵2n（n+1）（n+2）（n+3）+12=2（n2+3n）（n2+3n+2）+12，假设n2+3n+1=t，则t为奇数，故令t=2k+1，∴原式=4（2k2+2k+3）．若原式可表示为两个正整数x，

1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2)=(1+2+..+n)*n^2-(1^3+2^3+..+n^3)其中：1+2+3+..+n=n*(n+1)/21^3+2^3+

∵1*2*4+2*4*8+`````+n*2n*4n=1*2*4(1+2^3+...+n^3)1*3*6+2*6*12+````+n*3n*6n=1*3*6(1+2^3+...+n^3)∴(1*2*4

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^

先证明对于任意x≠0,1+xf(0)=1>0,即1+x

当n=1时,左边=4,右边=4,等式成立假设n=k时,1×2^2+2×3^2+3×4^2+...+k×(k+1)^2=k×(k+1)×(3k^2+11k+10)/12当n=k+1时,左边=1×2^2+

n=1,3^2=1/3(4+12+11)成立假设n=k时,3^2+...(2k+1)^2=k/3(4k^2+12k+11)则n=k+1时左边=3^2+...+(2k+1)^2+(2k+3)^2=k/3

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)(

(n+1)(n+2)/1+(n+2)(n+3)/1+(n+3)(n+4)/1=(n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)=(n+2)(n+1+n+3)+n^2+7n+12=(n+

a^(n+1)b^n-4a^(n+2)+3ab^n-12a^2=a^(n+1)(b^n-4a)+3a(b^n-4a)=(b^n-4a)[a^(n+1)+3a]=a(b^n-4a)(a^n+3)