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  • 数列求和常用公式

    1)1+2+3+......+n=n(n+1)÷2


    2)1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)÷6


    3) 1^3+2^3+3^3+......+n^3=( 1+2+3+......+n)^2
         =n^2*(n+1)^2÷4


    4) 1*2+2*3+3*4+......+n(n+1)
        =n(n+1)(n+2)÷3


    5) 1*2*3+2*3*4+3*4*5+......+n(n+1)(n+2)
        =n(n+1)(n+2)(n+3)÷4


    6) 1+3+6+10+15+......
       =1+(1+2)+(1+2+3)+(1+2+3+4)+......+(1+2+3+...+n)
       =[1*2+2*3+3*4+......+n(n+1)]/2=n(n+1)(n+2) ÷6


    7)1+2+4+7+11+......
      =1+(1+1)+(1+1+2)+(1+1+2+3)+......+(1+1+2+3+...+n)
      =(n+1)*1+[1*2+2*3+3*4+......+n(n+1)]/2
      =(n+1)+n(n+1)(n+2) ÷6


    8)1/2+1/2*3+1/3*4+......+1/n(n+1)
       =1-1/(n+1)=n÷(n+1)


    9)1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+......+1/(1+2+3+...+n)
       =2/2*3+2/3*4+2/4*5+......+2/n(n+1)
       =(n-1) ÷(n+1)


    10)1/1*2+2/2*3+3/2*3*4+......+(n-1)/2*3*4*...*n
      =(2*3*4*...*n- 1)/2*3*4*...*n


    11)1^2+3^2+5^2+..........(2n-1)^2=n(4n^2-1) ÷3


    12)1^3+3^3+5^3+..........(2n-1)^3=n^2(2n^2-1)


    13)1^4+2^4+3^4+..........+n^4
      =n(n+1)(2n+1)(3n^2+3n-1) ÷30


    14)1^5+2^5+3^5+..........+n^5
      =n^2 (n+1)^2 (2n^2+2n-1) ÷ 12


    15)1+2+2^2+2^3+......+2^n=2^(n+1) – 1

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  • 原文地址:https://www.cnblogs.com/Jonlee/p/149746.html
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