MI Question

Proof
Let P(N): T1*T2*T3*...*Tn = (n+2)/(2n+2)
when n=1,
L.H.S=T1
=1-1/(1+1)^2 (by Tn=1- 1/(n+1)^2)
=3/4
R.H.S=(1+2)/[2(1)+2]
=3/4
so P(1) true
assume P(k) true for all integers k=1 or K>1
i.e T1*T2*T3*...*Tk = (k+2)/(2k+2)
when n=k+1,
L.H.S=T1*T2*T3*...*T k*T(K+1)
=(k+2)/(2k+2)*T(K+1) (by assume)
=(k+2)/(2k+2)*[1- 1/(k+1+1)^2] (by Tn=1- 1/(n+1)^2)
=(k+2)/(2k+2)- (k+2)/(k+2)^2*(2k+2)
=(k+2)/(2k+2)-1/(k+2 )(2k+2)=(k+2)/2(k+1) -1/2(k+2)(k+1)
=[(k+2)^2-1]/(k+2)(2)(k+1)
=[(k+2+1)(k+2-1)]/(k+2)(2)(k+1), [Let A=k+2,B=1,A^2-B^2=(A+B)(A-B)]
=(k+3)(k+1)/(k+2)(2) (k+1)
=(k+3)/2(k+2)
=(K+3)/(2k+4)
=R.H.S
so P(K+1) true
By M.I,P(n) true for all positive integers n

when n=1
T1=1-(1+1)^-2=3/4=(1+2)/(2*1+2)
T1 is true.
T1*...Tn*T(n+1)=(n+2)/(2n+2)*(1-(n+2)^-2)=(n+3)/(2n+4)=((n+1)+2)/(2(n+1)+2)
when Tn is true, T(n+1) is true.
and T1 is true so ....
我ｄ格式，可能錯，我第一次做，唔好見怪