13-(2n-1)24
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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)/
证明:当k=1时1/2+1/3+1/4=13/12=26/24>25/24结论成立.假设k=n时结论成立,即1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24当k=n
增加了:1/(2k+1)+1/(2k+2)-1/(k+1)通分后,上式是大雨零的,所以成立
[n^2+(n+1)^2]/n(n+1)=n/(n+1)+(n+1)/n再问:我也化到那这步,还可以化吗?再答:=n/(n+1)+(n+1)/n=n/(n+1)+(1/n)+1很难说哪一步更好,因为这
证明:假设当n=k时,A=1/(k+1)+1/(k+2)+…+1/(k+k)>13/24成立,则当n=k+1时,左边=1/(k+2)+1/(k+3)+…+1/(k+1+k+1)=A+1/(k+1+k)
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
这道题可以是等差数列,公差为1可以写成0+1+2+3+.n=n(n+1)/2
是不是求证这个多项式能被13整除?N=(5^2)*(3^2n+1)*(2^n)-(3^n)*(6^n+2)=5^2*3^2n+1*2^n-3^n*(2*3)^n+2=5^2*3^2n+1*2^n-3^
根据题中所给式子,得f(n+1)-f(n)=1(n+1)+1+1(n+1)+2+1(n+1)+3+…+13(n+1)-(1n+1+1n+2+1n+3+…+13n)=13n+1+13n+2+13n+3-
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^
可以用归纳法比较容易首先,n=1比较容易证明然后假设n时成立求n+1时的式子,代入得到
先证明对于任意x≠0,1+xf(0)=1>0,即1+x
n(n+1)(n+2)=(n平方+n)(n+2)=n^3+3n^2+2n再答:望采纳!再答:不懂可以问我再问:啊咧,可以加你QQ么再问:3乘以27乘以9=3的x次方,则x等于多少?
设A=1*3*5*…*(2n-3)*(2n-1),则2*4*6*…*(2n-2)*(2n)A=(2n)!,(2^n)*1*2*3*…*n*A=(2n)!即(2n-1)!=(2n)!/[(2^n)*n!
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
1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n的每一项都>=1/2n,共有n个,所以1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n>n*1/2n=1/2,令m/24
证明:①当n=1时,左边=2,右边=13×1×2×3=2,等式成立;②假设当n=k时,等式成立,即1×2+2×3+3×4+…+k(k+1)=13k(k+1)(k+2)则当n=k+1时,左边=13k(k