anbn=1,an=n*2 3n 2
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n=1/an=1/(n^2+3n+2)=1/[(n+1)(n+2)]=1/(n+1)-1/(n+2)S10=b1+b2+...+b10=1/2-1/3+1/3-1/4+...+1/11-1/12=1/
1.an-an-1=2(n-1)-1=2(n-1)2n-2=-12n=2-12n=1n=1/22.3+(n-1)(-2)=-2n-53-2n+2=-2n-55=-5题目有错,无解.3.2+(n-1)x
因an=2^n,bn=2n-1所以anbn=(2n-1)2^n所以tn=a1b1+a2b2.+anbn=1*2+3*2^2+5*2^3+.+(2n-3)2^(n-1)+(2n-1)2^n两边乘以2得2
n=1/(n2+3n+2)=1/((n+1)(n+2))S10=1/(2*3)+1/(3*4).+1/(11*12)=1/2-1/3+1/3-.+1/11-1/12=1/2-1/12=5/12
an=(n+1)(n+2)anbn=1bn=1/an=1/[(n+1)(n+2)]=[(n+2)-(n+1)]/[(n+1)(n+2)]=(n+2)/[(n+1)(n+2)]-(n+1)/[(n+1)
an=n^2+3n+2=(n+1)(n+2)bn=1/[(n+1)(n+2)]=1/(n+1)-1/(n+2)S10=b1+b2+..+b10=(1/2-1/3)+(1/3-1/4)+..+(1/11
不知道你的题目是不是这样
(1)设数列{an}的公差为d,数列{bn}的公比为q,则由题意知a1b1=1(a1+d)(b1q) =4(a1+2d)(b1q2) =12 ,因为数列{an}各项为正数
a_(n+1)=(1+1/(n+1))^(n+1)=(1/n+1/n+...+1/n+1/(n+1))^(n+1)>[(n+1)(1/((n^n*(n+1)))开(n+1)次方根]^(n+1)(均值不
∵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)
令Tn为{anbn}的前n项和,那么:Tn=a1b1+a2b2+…+anbn=1×20+3×21+5×22+…+(2n-1)•2n-12Tn=1×21+3×22+5×23+…(2n-1)•2n∴Tn=
C(k,n)ak=n!/((n-k)!*k!)*(k(k+1))/2=(n-1)!/((n-k)!(k-1)!)*(n(k+1))/2=C(k-1,n-1)*n/2*(k+1)An=n/2*[C(0,
n是(1/2)n还是1/(2n)
cn=anbn=(3n-1)*2^nSn=2*2^1+5*2^2+……+(3n-1)*2^n2Sn=2*2^2+……+(3n-4)*2^n+(3n-1)*2^(n+1)相减:Sn=(3n-1)*2^(
等我算算啊,几分钟
(n^2+n)x^2-(2n+1)x+1=0由根与系数的关系x1+x2=(2n+1)/(n^2+n)x1x2=1/(n^2+n)|AnBn|=|x1-x2|=√(x1-x2)^2=√[(x1+x2)^
(1)证明:∵在数列{a[n]}中,已知a[n]+a[n+1]=2n(n∈N*)∴用待定系数法,有:a[n+1]+x(n+1)+y=-(a[n]+xn+y)∵-2x=2,-x-2y=0∴x=-1,y=
∵数列{an}、{bn}是等差数列,且其前n项和分别为An、Bn,由等差数列的性质得,A21=(a1+a21)×212=21a11,B21=(b1+b21)×212=21b11,∵足AnBn=7n+1
因为Sn=2^n-1所以S(n-1)=2^(n-1)-1所以an=Sn-S(n-1)=2^(n-1)(n>=2)因为S1=a1=2^1-1=1=2^0所以an=2^(n-1)(n>=2)因为bn=n所