展开全部
a1+2a2+3a3+……+nan=(n+1)/2*an+1
a1+2a2+3a3+……+(n-1)an-1=nan/2
两式相减nan=(n+1)/2*an+1-nan/2
所以3nan=(n+1)an+1
设bn=nan
那么b1=1,3bn=bn+1
bn=3^(n-1)
an=3^(n-1)/n
a1+2a2+3a3+……+(n-1)an-1=nan/2
两式相减nan=(n+1)/2*an+1-nan/2
所以3nan=(n+1)an+1
设bn=nan
那么b1=1,3bn=bn+1
bn=3^(n-1)
an=3^(n-1)/n
展开全部
Sn = (n+1)/2 a(n+1) ——①
S(n-1) = n/2 an ——②
①-②
an = [n+1]/2 a[n+1] - n/2 an
[n+2]/2 an = [n+1]/2 a[n+1]
a[n+1] / an =[ n+2] /[ n+1]
用累乘法:
a2 / a1 = 3/2
a3 / a2 = 4/3
a4 / a3 = 5/4
︰
an / a[n-1] =[ n+1 ]/ n
∴ an / a1 = 3/2 × 4/3 × 5/4 × … × n+1 / n = [n+1] /2
∵ a1 = 1
∴ an =[ n+1] /2
S(n-1) = n/2 an ——②
①-②
an = [n+1]/2 a[n+1] - n/2 an
[n+2]/2 an = [n+1]/2 a[n+1]
a[n+1] / an =[ n+2] /[ n+1]
用累乘法:
a2 / a1 = 3/2
a3 / a2 = 4/3
a4 / a3 = 5/4
︰
an / a[n-1] =[ n+1 ]/ n
∴ an / a1 = 3/2 × 4/3 × 5/4 × … × n+1 / n = [n+1] /2
∵ a1 = 1
∴ an =[ n+1] /2
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