Sn Tn =(2n-3) 4n-3) (a1 a15) 2(b3 b9)

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=

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

令g(n)=f(n)/(n-1)!,h(n)=g(n)/n=f(n)/n!那么g(n)=g(n-2)+h(n-3)+h(n-4)对n求和可得g(n)=1+h(1)+h(2)+...+h(n-3)因此g

∵anbn＝2an2bn＝a1+a2n−1b1+b2n−1=(2n−1)(a1+a2n−1) 2(2n−1)(b1+b2n−1) 2＝s2n−1T2n−1∴anbn=2(2n−1)

c2^n(n-1)

1)假设当自然数n>=4时,n^3>3n^2+3n+1当n=4时,4^3=64>3*4^2+3*4+1=61令n=k时,k^3>3k^2+3k+1成立,k>=4则n=k+1时,(k+1)^3=k^3+

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

1)C(n,0)+2C(n,1)+3C(n,2)+4C(n,3)+...+(n+1)C(n,n)=C(n,0)+2C(n,1)+3C(n,2)+4C(n,3)+...+(n+1)C(n,n)-（C(n

［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~~~└

设等差数列{an}和{bn}的公差分别为d1 和d2，则由题意可得S1T1＝a1b1＝2×13×1+1=12，即2a1=b1．再由S2T2＝a1+a2b1+b2＝2a1+d12b1+d2＝2

由等差数列的性质和求和公式可得：a9b5+b7+a3b8+b4=a9b1+b11+a3b1+b11=a3+a9b1+b11=a1+a11b1+b11=11(a1+a11)211(b1+b11)2=S1

M=(N-1)×1+(N-2)×2+(N-3)×4+(N-4)×8+(N-5)×16+(N-6)×32+(N-7)×64+...（N-n-1)×2n……①2M=(N-1)×2+(N-2)×4+(N-3

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

∵等差数列{an}、{bn}，∴an=a1+a2n−12，bn=b1+b2n−12，∴anbn=nannbn=n(a1+a2n−1)2n(b1+b2n−1)2=S2n−1T2n−1，又SnTn=7n+

1/1*2*3+1/2*3*4+……1/n(n+1)(n+2)=1/2(1/1*2-1/2*3)+1/2(1/2*3-1/3*4)+...+1/2[1/n(n+1)-1/(n+1)(n+2)]=1/2

(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+

过程较繁琐,但是道理很清晰1/n(n+1)(n+2)=1/2n(n+1)-1/2(n+1)(n+2)=1/2(1/n+1/(n+2)-2/(n+1))利用数学归纳法先证n=1成立设n=k成立,证明n=