数列满足a1 2a2 2^3a3 -- 2^n-1an=n 2,则an=?

设公比为q2a1+a3=3a22a1+a1q²=3a1qq²-3q+2=0(q-1)(q-2)=0q=1或q=2a3+2是a2、a4的等差中项2(a3+2)=a2+a42(a1q&

因为数列｛an｝满足a1,a2-a1,a3-a2...是以1为首相,3为公比的等比数列所以a1=1,an-a(n-1)=1*3^(n-1)=3^(n-1)(n≥2)由叠加法有：an=a1+(a2-a1

a1+2a2+3a3+……+(n-1)a(n-1)+nan=2^na1+2a2+3a3+……+(n-1)a(n-1)=2^(n-1)两式相减得nan=2^n-2^(n-1)nan=2^(n-1)an=

1,1/a1+2/a2+3/a3+…+n/an=2n那么1/a1+2/a2+3/a3+…+(n-1)/a(n-1)=2(n-1)两式相减,得：n/an=2n-2(n-1)=2那么an=n/22,Sn=

取n和n-1,两等式互减再答：n等于1另算

我告诉你方法吧!通过a3-a2-(a2-a1)求出d=1然后再根据an+1-an=a2-a1+(n-1)d,求出an+1-an,再将an+1-an,an-an-1…a2-a1进行叠加,即可求到an,同

an=2^n步骤：等比数列{an},=>an=a1*q^(n-1),(a1、q不为0)=>a2=a1q,a3=a1q^2,a4=a1q^3,2a1+a3=3a2=>2a1+a1q^2=3a1q,=>q

A1+3A2+3²A3++3^（n-1）An+3^n*A(n+1)=(n+1)/3下减上：3^n*A(n+1)=1/3A(n+1)=3^(-n-1)则通项An=3^(-n)

1.设Qn=n/3Qn+1=(n+1)/3Qn+1-Qn=3^n*an+1=1/3an+1=1/3^(n+1)an=1/3^n2.bn=n*3^n

令n=1时,a1=1*2*3=6；依题意：a1+2a2+3a3+.+nan=n(n+1)(n+2),a1+2a2+3a3+.+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)两式相减,得

a1+a2+a3+a4+a5=a1+q*a1+q^2*a1+q^3*a1+a^4*a1=a1(1+q+q^2+q^3+q^4)=a1(1-q^5)/(1-q)=3——(1)a1²+a2&su

(1)a1+3a2+…+3^(n-2)an-1=(n-1)/3a1+3a2+…+3^(n-1)an=(n-1)/3+3^(n-1)an=n/3an=(1/3)^n.(2)bn=n/an=n3^nSn=

1、①A1+3A2+3^2*A3+...+3^(n-1)*An=n/3,又A1+3A2+3^2*A3+...+3^(n-)*An-1=（n-1）/3,（比已知的式子最后少写一项,即有n-1项）,两式相

a1＋3a2＋3²a3＋…＋3^(n－1)an=n/3a1＋3a2＋3²a3＋…＋3^(n－2)a(n－1)=(n－1)/3=n/3-1/3（n≥2）两式相减得：3^(n-1)an

∵数列{a[n]}满足a[1]+2a[2]+3a[3]+...+na[n]=(n+1)(n+2)∴a[1]+2a[2]+3a[3]+...+na[n]+(n+1)a[n+1]=(n+2)(n+3)将上

记Tn表示{an}的前n项和a1^3+a2^3+a3^3+...+an^3=(a1+a2+a3+...+an)^2……(1)a1^3+a2^3+a3^3+...+a^3(n-1)=(a1+a2+a3+

an满足an满足a1+2a2+3a3+...+nan=2^n所以有a1+2a2+3a3+...+（n-1）a（n-1）=2^（n-1）上面两式作减法有nan=2^n-2^(n-1)=2^(n-1)an

由a1+3a2+3^2a3+……+3^(n-1)an=n/3和a1+3a2+3^2a3+……+3^(n-1)an+3^na_(n+1)=(n+1)/3得3^n*a_(n+1)=1/3所以a_(n+1)

证明：因为：a1+2a2+3a3+…+nan=n(n+1)(n+2),记:bn=nan,那么：b1+b2+...+bn=n(n+1)(n+2)将n-1带入,得：b1+b2+...+b(n-1)=(n-

1、依题a1=1-a1得出a1=0.5a1+a2=2-a2得出a2=0.75a1+a2+a3=3-a3得出a3=0.8752、设p=n-1显见∑(an-1)=-an∑(ap-1)=-ap∑(an-1)