N^4-3N^2-9
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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+
n-1=再问:29/20n-4n=
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=
#includevoidmain(){inti;doublen,s=0;doubelt=1;printf("请输入N:\n");scanf("%lf",&n);for(i=1;i
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/
利用(1+1/n)^n在n趋于无穷极限为e.构造[1+(-6)/(3n^2+4)]^[(3n^2+4)/(-6)]形式.结果为e^(-2)
n是趋于无穷大么?就按这个解答.分子分母同除以n^4,化为[1/n*(1+1/n)(1+2/n)(1+3/n)]/(1+1/n^2+1/n^4),由于n趋于无穷大,所以1/n、2/n、3/n、1/n^
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+
(2n+3n)的2次方-(6n+9n)原式=5n的2次方-15n=25n-15n=10n
[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~~~└
分子=8*(1^3+2^3+……+n^3)分母=27*(1^3+2^3……+n^3=>原式=(8/27)^2=64/729
lim(n->∞)[(n^2+3n-8)/(4n^2+2n+3)]=lim(n->∞)[(1+3(1/n)-8(1/n^2))/(4+2(1/n)+3(1/n^2))]=1/4
16n^4+4n^3+6n^2+7n=0n(16n^3+4n^2+6n+7)=0n=016n^3+4n^2+6n+7=0(无实数解)所以原方程的解是n=0
先证明对于任意x≠0,1+xf(0)=1>0,即1+x
原式=lim(n->∞)[2+1/n]/[1+1/(n^2)+4/(n^3)](分子分母同除以n^3)=lim(n->∞)[2+0]/[1+0+0](n在分母的项都趋于0)=lim(n->∞)2=2
数列1+4+…+3n-1的和Sn=n+3n(n-1)/2=n+3n/2-3n/2=3n/2-n/2lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2)=lim(3n^2-n/2n^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+
不可能吧!当n=1时,原式=1x2x3x4x5=120当n=2时,原式=2x3x4x5x6=720都不是完全平方数再问:没错,后来才发现,老师题目出错了。应为:n(n+1)(n+2)(n+3)+1还是