已知数列bn=9n+4/2*4n,求数列bn的前n项和
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已知数列bn=9n+4/2*4n,求数列bn的前n项和
已知数列bn=9n+4/2*4n,求数列bn的前n项和
已知数列bn=9n+4/2*4n,求数列bn的前n项和
把4/2^(4n)化为4/4^(2n)=4^(1-2n),然后9n与4^(1-2n)(错位相减）分别求和再加起来即可
bn = (9n+4)/2^(4n)
= 9[ n.(1/16)^n] + 2^(2-4n)
summation bn
= summation {9[ n.(1/16)^n] + 2^(2-4n)}
=( summation {9[ n.(1/16)^n] } ) + 2^(-2)( 2^(-4n) -1 ) / [2^(-4) -1]
=...
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bn = (9n+4)/2^(4n)
= 9[ n.(1/16)^n] + 2^(2-4n)
summation bn
= summation {9[ n.(1/16)^n] + 2^(2-4n)}
=( summation {9[ n.(1/16)^n] } ) + 2^(-2)( 2^(-4n) -1 ) / [2^(-4) -1]
= ( summation {9[ n.(1/16)^n] } ) + (4/15)( 2^(-4n) -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
multiply both side by x
x+2x^2+..+nx^n = x{nx^(n+1) -(n+1)x^n +1 } /(x-1)^2
put x=1/16
1(1/16)+2(1/16)^2+...n(1/16)^n
= (16/225) [ n (1/16)^(n+1) -(n+1)(1/16)^n+1 ]
summation bn
= ( summation {9[ n.(1/16)^n] } ) + (4/15)( 2^(-4n) -1 )
= (16/25)[ n (1/16)^(n+1) -(n+1)(1/16)^n+1 ] + (4/15)( 2^(-4n) -1 )
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