My method: S=a1*n+[n(n-1)/2]*d d=1 S=52 so: a1*n+n(n-1)/2=52 then: n^2+(2a1-1)n-104=0 104=2*2*2*13 so n must=8, becasue only when (n+13)*(n-8)=0, then n=8
My method: S=a1*n+[n(n-1)/2]*d d=1 S=52 so: a1*n+n(n-1)/2=52 then: n^2+(2a1-1)n-104=0 104=2*2*2*13 so n must=8, becasue only when (n+13)*(n-8)=0, then n=8
My method: S=a1*n+[n(n-1)/2]*d d=1 S=52 so: a1*n+n(n-1)/2=52 then: n^2+(2a1-1)n-104=0 104=2*2*2*13 so n must=8, becasue only when (n+13)*(n-8)=0, then n=8
-- by 会员 mjshan33 (2012/2/8 19:55:08)
好方法!!
-- by 会员 anikakoko (2012/2/8 20:24:10)
xie xie lou zhu, ni xin ku le!!!!! wo dian nao shu ru fa you huai le ,zhi neng da pin yin huo zhe ying yu....