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n+1 1+An

(1) a(n+1)=an/(an+1) 1/a(n+1) = (an+1)/an 1/a(n+1) -1/an = 1 =>(1/an)是等差数列 1/an -1/a1= n-1 1/an =n an =1/n (2) bn =1/(2^n.an) = (1/2)[n(1/2)^(n-1)] consider 1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1) 1+2x+..+nx^(n-1) =[(x^(n+1)- ...

a(n+1)=an+1/n-1/(n+1) a(n+1)+1/(n+1)=an+1/n 设bn=an+1/n 则:b(n+1)=bn b1=a1+1/1=2 所以:数列bn为常数列bn=2 所以an+1/n=2 an=2-1/n

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...

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a1=1, a2=2, a3=5, a4=26, a5=26*26+1 没有通项公式,只能一个一个顺着带入计算,既不是等差,也不是等比

您好:如题,一个万能法: ∵f(n)=a(n)+a(n+1) ∴f(n-1)=a(n-1)+a(n-1+1)=a(n-1)+a(n) 两式相减得f(n)-f(n-1)=a(n+1)-a(n-1) 而f(n)-f(n-1)=a(n) ∴有a(n)=a(n+1)-a(n-1) 即a(n+1)=a(n)+a(n-1) 即a(n)=a(n-1)+a(n-2) 而f(1)=a1+a2=a1 ∴ a2=0 同理得...

#includeint main(){int a=1,b=4,c,s=0,k=0; for(;a+b=1000) {if(k)printf("+"); else k=1; printf("%d",c); s+=c; } } printf("=%d\n",s); return 0;}

裂项相减 1/[n(n+1)]=1/n-1/(n+1)

a(n+1)/an=n/n+1 则an/a(n-1)=(n-1)/n a(n-1)/a(n-2)=(n-2)/(n-1) .......... a2/a1=1/2 叠乘 an/a1=1/n 因a1=1 故an=1/n

见过一个类似题目,供参考: 数列{an}中,a1=1/2, a(n+1)=an^2+an, 求证:1/(a1+1)+1/(a2+1)+......+1/(an+1)

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