bn等于2的an次方_

an=Sn-S(n-1)=2^n-2^(n-1)=2^(n-1)所以bn=n-1所以Tn=0²-1²+2²-3²+4²+(-1)^(n-1)*(n-1

错位相减法Sn=1×2+3×2²+5×2³+.+(2n-1)×2ⁿ①2Sn=2²+3×2³+5×2⁴+.+(2n-3)×2ⁿ

不好意思,你的题应该出错了应该是An+1=2An+2^n(^代表2的次方）这样算来对第一题：证明只需证明Bn+1-Bn为常数即可.证明：Bn+1-Bn=An+1/2^(n+1)-An/2^n=(An+

因为an=2n所以bn=2n×3的n次方∴Sn=2*3+2×2*3＾2+2*3*3＾3+……+2*n*3＾n两边同时除以21/2Sn=3+2*3＾2+……+n*3＾n⑴3/2Sn=3＾2+2*3＾3+

∵an•bn=1∴bn=1n2+3n+2=1(n+1)(n+2)∴s10=12×3+13×4+   + 110×11+111×12=（12-13）+(13−14

/>a(n+1)=Sn+2ⁿS(n+1)-Sn=Sn+2ⁿS(n+1)-2Sn=2ⁿ等式两边同除以2ⁿS(n+1)/2ⁿ-Sn/2^(n-1

从题意可知,bn=n*2^(2n-1)sn=b1+b2+b3+.bn=1*2+2*2^3+3*2^5+……+n*2^(2n-1)左右乘以2^2得：4*sn=1*2^3+2*2^5+3*2^7……+（n

(1)a(n+1)=2an+2^(n+1)等式两边同除以2^(n+1)a(n+1)/2^(n+1)=an/2ⁿ+1a(n+1)/2^(n+1)-an/2ⁿ=1,为定值a1/2=

d=(21-9)/3=4a1=5an=a1+(n-1)*d=4n+1bn=2^(4n+1)bn-1=2^[4(n-1)+1]=2^(4n-3)bn+1=2^[4(n+1)+1]=2^(4n+5)bn+

证：由数列{an}是等差数列,得an=a1+(n-1)d,其中a1为首项,d为公差.b1b2b3=[(1/2)^(a1)][(1/2)^(a1+d)][(1/2)^(a1+2d)]=(1/2)(a1+

看图咯

数列an的前n项和Sn=n^2-2nn=1时a1=S1=1-2=-1n>=2时an=Sn-S(n-1)=(n^2-2n)-((n-1)^2-2(n-1))=2n-3n=1时,满足an=2n-3∴an=

先证明bn=b^n/2^n=(b/2)^n(1)bn-1=(b/2)^(n-1)(2)(1)÷(2)bn/bn-1=b/2,是定值所以bn是等比数列计算anan=2an-1+2^(n+1)an=2an

n=an/2^(n-1)得an=bn*2^(n-1)a(n-1)=b(n-1)*2^(n-2)由an=2a(n-1)+2^(n-1),得bn*2^(n-1)=2*b(n-1)*2^(n-2)+2^(n

由题意可得：(x-an)(x-an+1)=x2-bnx+2化简后的x2-(an+an+1)x+an*an+1=x2-bnx+2所以可得：bn=an+an+1an*an+1=2所以可得：a1a2=2(1

∵数列{an}是等差数列,∴an-a(n-1)=d∵bn/b(n-1)=2^an/[2^a(n-1)]=2^[an-a(n-1)]=2^d∴{bn}是等比数列,公比为2^d

Sn=3n-2,B1=S1=1Bn=Sn-S(n-1)=3(n>1)根据Bn=3的n次方乘An,当n=1时,A1=1/3当n>1时,An=3的n-1次幂再问：为什么Bn=Sn-S(n-1)=3再答：S

∵An,A(n+1)是方程x^2-(2n+1)x+1/Bn=0的两个根∴An+A(n+1)=2n+1,An*A(n+1)=1/Bn（根与系数的关系）∴Bn=1/An*A(n+1)这题我看过,应该是缺了

/>1,A(n+1)=2An+2^n,两边除以2^n得A(n+1)/2^n=An/2^(n-1)+1,即B(n+1)=Bn+1,Bn是等差数列.2,B1=A1=1,则Bn=n,即An=n2^(n-1)