关于数列求和.
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关于数列求和.
关于数列求和.
关于数列求和.
(1)
(a-1)+(a^2-2)+...+(a^n-n)
=(a+a^2+...+a^n) -(1+2+3+...+n)
=a(a^n-1)/(a-1) - n(n+1)/2
(2)
1+3(3)+5(3)^2+...+(2n-1).3^(n-1)
an = (2n-1).3^(n-1)
= 2.[n.3^(n-1)] - 3^(n-1)
1+3(3)+5(3)^2+...+(2n-1).3^(n-1)
=a1+a2+...+an
=2[ ∑(i:1->n) i.(3)^(i-1) ] - (1/2)(3^n-1)
consider
1+x+x^2+...+x^n = (x^(n+1) -1)/(x-1)
1+2x+...+nx^(n-1) = [(x^(n+1) -1)/(x-1)]'
= (nx^(n+1) -(n+1)x^n +1)/(x-1)^2
put x=3
∑(i:1->n) i.(3)^(i-1) = (1/4)[ n.(3)^(n+1) -(n+1).(3)^n +1 ]
= (1/4) [ 1+ (2n-1).(3)^n ]
1+3(3)+5(3)^2+...+(2n-1).3^(n-1)
=a1+a2+...+an
=2[ ∑(i:1->n) i.(3)^(i-1) ] - (1/2)(3^n-1)
=(1/2)[ 1+ (2n-1).(3)^n ] -(1/2)(3^n-1)
= 1 + (n-1)3^n
(3)
1/[(3n-2)(3n+1)] = (1/3)[ 1/(3n-2) - 1/(3n+1) ]
1/(1x4) + 1/(4x7)+...+1/[(3n-2)(3n+1)]
= (1/3)[ 1- 1/(3n+1)]
(4)
1/[√n+ √(n+1)]
= √(n+1) -√n
1/[√1+ √2] +1/[√2+ √3] +...+1/[√n+ √(n+1)]
=√(n+1) - 1