已知1+2+3+.+n/1+3+5+.+(2n-1)=10/19.求nrtttttttttttttttttt

Question

已知1+2+3+.+n/1+3+5+.+(2n-1)=10/19.求n
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Answer 1

由等差数列求和公式
1+2+3+.+n=n(n+1)/2
1+3+5+.+(2n-1)=n(1+2n-1)/2=2n^2/2=n^2
(1+2+3+.+n)/(1+3+5+.+(2n-1))=10/19
即[n(n+1)/2]/n^2=10/19
(n+1)/2n=10/19
20n=19n+19
n=19