设边长为1的正三角形ABC的边BC上有n-1个等分点,沿点B到点C的方向,依次为P1,P2,…Pn-1,若Sn=AB*A
设边长为1的正三角形ABC的边BC上有n-1个等分点,沿点B到点C的方向,依次为P1,P2,…Pn-1,若Sn=AB*AP1+AP1*AP2+APn-1*AC(以上均为向量)求证:Sn=5n的平方-2/6n
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APk*AP(k+1)=（AB+kBP1）（AB+(k+1)BP1）=
AB*AB+(2k+1）BP1*AB+k*(k+1)BP1*BP1=
1-(2k+1)/2n+(k*k+k)/n*n
Sn=AB*AP1+(1+1..+1)(n-1个）-
[3/2n+5/2n+..2n-1/2n]+[(1+1)/n*n+..(n*n-n)/n*n]
=1-1/2n+n-1-(2n+2)(n-1)/4n+
[n(n-1)/2+n(n-1)(2n-1)/6]/n*n]
=n-n/2+1/2-1/2n+n/3-1/2+1/6n
=5n/6-1/3n
=(5n*n-2)/6
AB*AB+(2k+1）BP1*AB+k*(k+1)BP1*BP1=
1-(2k+1)/2n+(k*k+k)/n*n
Sn=AB*AP1+(1+1..+1)(n-1个）-
[3/2n+5/2n+..2n-1/2n]+[(1+1)/n*n+..(n*n-n)/n*n]
=1-1/2n+n-1-(2n+2)(n-1)/4n+
[n(n-1)/2+n(n-1)(2n-1)/6]/n*n]
=n-n/2+1/2-1/2n+n/3-1/2+1/6n
=5n/6-1/3n
=(5n*n-2)/6
1年前
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