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已知A1=1An+1%An=n+1求数列An分之一的前十项和

解: a(n+1)-an=n+1=½[(n+1)²-n²]+½ [a(n+1)-½(n+1)²]-(an-½n²)=½,为定值 a1-½×1²=1-½=½ 数列{an-½n²}是以½为首项,½为公差的等差数列 an-½n...

裂项

a(n+1)=an+2^n a(n+1)-an=2^n an-a(n-1)=2^(n-1) ......... a3-a2=2^2 a2-a1=2^1 以上等式相加得 a(n+1)-a1=2^1+2^2+...+2^n a(n+1)-a1=2*[1-2^n]/(1-2) a(n+1)-a1=2^(n+1)-2 a(n+1)-2=2^(n+1)-2 a(n+1)=2^(n+1) an=2^n sn=a1+a2+....+an =2^1+2...

解: (1) n≥2时, 2an=2Sn-2S(n-1)=(n+1)an-na(n-1) (n-1)an=na(n-1) an/n=a(n-1)/(n-1) a1/1=2/1=2,数列{an/n}是各项均为2的常数数列 an/n=2 an=2n n=1时,a1=2×1=2,同样满足表达式 数列{an}的通项公式为an=2n (2) 4/[an(an+2)]=4/[2n×(2n+2...

由题意可得知, 不懂步骤的,详细可以再问

(Ⅰ)由a1=2,an+1=2an,得an=2n(n∈N∗). 由题意知,当n=1时,b1=b2-1,故b2=2, 当n≥2时,b1+12b2+13b3+…+1n−1bn−1=bn-1,和原递推式作差得, 1nbn=bn+1−bn,整理得:bn+1n+1=bnn, ∴bn=n(n∈N∗); (Ⅱ)由(Ⅰ)...

第一步:由已知条件Sn=1/2(n+1)(an+1)-1,可知: ①Sn-S(n-1)=a(n)=[1/2(n+1)(an+1)-1]-{(1/2)*n*[a(n-1)+1]-1} ②S(n-1)-S(n-2)=a(n-1)=(1/2)*n*[(a(n-1)+1]-1/2*(n -1)*[a(n-2)+1] 由①式可得:a(n)=(n+1)*a(n)/2+(n...

an/an-1=(n+1)/(n-1) an/a₁=(n+1)/(n-1) * n/(n-2)*...*4/2 * 3/1=(n+1)n / 2*1 an=(n+1)n/2

两边同时加Sn Sn+1=(2+n)Sn/n+1/3n^2+n+2/3 根据一阶线性变系数差分方程的公式,该方程的通解为 Sn=[求和0到n-1(2x^2/3(x+1)(x+2)+2x/(x+1)(x+2)+4/3(x+1)(x+2))]*n(n+1)/2+Cn(n+1)/2 2x^2/3(x+1)(x+2)+2x/(x+1)(x+2)+4/3(x+1)(x+2)=2/3-(6x+4...

当n大于等于2时 S(n-1) = (n/3+1/3)*a(n-1) 把Sn - S(n-1)得 an = (n/3+2/3)*an - (n/3+1/3)*a(n-1) 整理得 an/a(n-1) = (n+1)/(n-1) a2/a1 * a3/a2 * a4/a3 * ... * a(n-1)/a(n-2) * an/a(n-1) an/a1 = 3/1 * 4/2 * 5/3 * ... * n/(n-2) * (n+1...

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