N4 L.N.PE BYJ-3*2.5 SC15-FC是什么意思

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)/

裂项相消法1/3【1/n-1/(n+3)+1/(n+3)-1/(n+6)+1/(n+6)-1/(n+9)】=1/(2n+18)1/3{1/n-1/(n+9)}==1/(2n+18)交叉相乘6n+54=

1/n(n+1)+1/(n+1)(n+2)＋1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/(n+99)(n+100)=1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/

(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

-n(n-4)^2

n/(n^2+1)+n/(n^2+2)+n/(n^2+3)+……+n/(n^2+n)n/(n²+n)+n/(²+n)+.+n/(n²+n)=n*n/(n²+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

这个就是二项式定理的逆用1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=1*C（n,0)+2C(n,1)+4C(n,2)+...+2^nC(n,n)=（1+2）^n=3^n明教为您解答

证明：（1）当n=1时，左边=1×2×3=6，右边=1×2×3×44＝6=左边，∴等式成立．（2）设当n=k（k∈N*）时，等式成立，即1×2×3+2×3×4+…+k×(k+1)×(k+2)＝k(k+

［n/2]+[n/3]+[n/4]+[n/5]+[n/6]=(30n+20n+15n+12n+10n)/60=87n/60=29n/60题目是不是打错了..等于29吧?这样n=60再问：是69~~~└

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^

只能大致写一下思路,具体计算你自己算吧.1、f(x)=求和（n＝3到无穷）x^n/n,f'(x)=求和(n=3到无穷)x^(n－1)=x^2/(1－x),因此f(x)=－0.5x^2－x－ln(1－x

先证明对于任意x≠0,1+xf(0)=1>0,即1+x

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+