Sn=(3n+1)/2-(n/2)an
当n=1时,a1=4/3=1+1/3=1+1/[1*(1+2)]
当n=2时,a2=13/12=1+1/[2*(1+2+3)
当n=3时,a3=31/30=1+1/[3*(1+2+3+4)
因此,可以猜想,an=1+2/[n(n+1)(n+2)]
然后再用数学归纳法证明,因为方法很机械,但又太烦了,而且你也不需要详细,我就省略了.
an-1=1+1/[n(n+1)(n+2)]-1=2/[n(n+1)(n+2)]
1/(an-1)=n(n+1)(n+2)/2
设数列bn=1/(an-1)＝n(n+1)(n+2)/2
n(n+1)(n+2) =1/2*[3!*C(n+2,3)]
Sbn =1/2*[3!*C(n+2,3)]
Sbn=1/2*{3!*[ C(3,3)+C(4,3)+C(5,3)+……+C(n+2,3)] }
=3!*C(n+3,4) /2
=3!(n+3)!/4!(n-1)!/2
=(n+3)(n+2)(n+1)n/8
tanAtanB=tanAtanC+tanBtanC
sinA/cosA * sinB/cosB =sinA/cosA * sinC/cosC + sinB/cosB * sinC/cosC
sinAsinBcosC=sinAsinCcosB+sinBsinCcosA
根据正弦定理,a/sinA=b/sinB=c/sinC=2R
代入,得
sinA=a/(2R),sinB=b/(2R),sinC=c/(2R)
abcosC/(4R^2)=accosB/(4R^2)+bccosA/(4R^2)
abcosC=accosB+bccosA
根据余弦定理,可得
cosC=(a^2+b^2-c^2)/(2ab)
cosA=(b^2+c^2-a^2)/(2bc)
cosB=(a^2+c^2-b^2)/(2ac)
代入,得
ab(a^2+b^2-c^2)/(2ab)=ac(a^2+c^2-b^2)/(2ac)+bc(b^2+c^2-a^2)/(2bc)
a^2+b^2-c^2=a^2+c^2-b^2+b^2+c^2-a^2
a^2+b^2=3c^2
(a^2+b^2)/c^2=3