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数列{an}中,a1=2,an+1=an+1/2^n,则an=

a(n+1)-an=2^nan-a(n-1)=2^(n-1)……a2-a1=2累加得a(n+1)-a1=2+2+2+……+2^na(n+1)=2^(n+1)an=2^n

a2=a1 + 2a3=a2 + 3an=an-1 + n左右两边加起来(a2+a3++an)=(a1+a2+.an-1)+2+3+4++n消去括号中相同项得到an=a1 + (n+2)(n-1)/2对于a1要代入验证

an =a(n-1)+1/2^(n-1)a(n-1)=a(n-2)+1/2^(n-2)……a3 =a2+1/2^2a2 =a1+1/2上面n-1个式子相加,得an=a1+1/2+1/2^2+1/2^3+……+1/2^(n-1) =2+0.5[1-1/2^(n-1)]/(1-1/2) =2+1-1/2^(n-1)=3-1/2^(n-1)所以选D

∵a1=2,an+1=an+n+1∴an=an-1+(n-1)+1,an-1=an-2+(n-2)+1,an-2=an-3+(n-3)+1,…,a3=a2+2+1,a2=a1+1+1,a1=2=1+1将以上各式相加得:an=[(n-1)+(n-2)+(n-3)+…+2+1]+n+1=(n1)[(n1)+1]2+n+1=(n

b-bn=a/2^n-an/2=2an+2^n/2^n-2an/2^n=1 所以{bn}是等差数列

1已知数列an满足an=2an-1+2^n-1(n>=2),有an-1=2(an-1-1)+2^n,两边同时除以2^n,得bn=bn-1+1故数列{bn}为首项b1=2,d=1的等差数列2由一问可知,an=(n+1)2^n+1故sn=n*(n+1)/2 +2*2+3*2^2+……+(n+1)*2^n用错位相减法

an+1=an+n+1 那么 an -a(n-1)=n a(n-1)-a(n-2) =n-1 . a3-a2 =3 a2-a1 =2 a1=2 累加an =(n+n+2)/2

解:由于 a(n+1)=an+2^(n+1) 则:左右同时除以2^(n+1)得:[a(n+1)/2^(n+1)]=[an/2^(n+1)]+1 [a(n+1)/2^(n+1)]=(1/2)[an/2^n]+1 设:bn=an/2^n 则:b(n+1)=(1/2)bn+1 则有:[b(n+1)-2]=(1/2)[bn-2] 则:[b(n+1)-2]/[bn-2]=1/2 则:{bn-2}为公比为1/2的等比数列 则:bn-2=(b1-2)*(1/2)^(n-1) =(1/2)^(n-1) 则:bn=an/2^n=(1/2)^(n-1)+2 则:an=2+2^(n+1) 则:an-2^n=2+2^n 不为等差数列 题目有误

an+1=an+(2的n次方) a(n+1)-2^(n+1)=an-2^n 所以:{an-2^n}为常数列!an-2^n=a1-2=0 所以:an=2^n

an=an-1+n-1+1an-1=an-2+n-2+1……a2=a1+1+1所有的加起来an=a1+(n-1)+(n-2)+……+2+1+(n-1)*1an=(n^2+n)/2

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