Prove T1+T2+...+Tn= (n+1)! -1

Let P(n) be T1+T2+......+Tn= (n+1)! -1 is true for all positive integers n.
When n=1
LHS=T1=1
RHS=2!-1=1=LHS
∴P(1) is true
Assume that P(k) is true where k is a positive integer
i.e. T1+T2+......+Tk= (k+1)! -1
For n=k+1
LHS=T1+T2+......+Tk+T(k+1)
=(k+1)!-1+(k+1)(k+1)!
=(k+1)![(1+k+1)]-1
=(k+1)!(k+2)-1
=(k+2)!-1
=RHS
∴P(k+1) is true
By the principle of MI, P(n) is true for all positive nos. n

2008-08-07 15:03:07 補充：
回001
每個人教其他人做功課的方法也不同，你為何要自命清高的認為你那種才是「教」而我的那種就不是呢？

For n=1,
L.H.S.= T1
=1(1!)
=1
R.H.S.= (1+1)! -1
=1
=L.H.S.
∴The statement is true for n=1.
Assume_T1+T2+...+Tk= (k+1)! -1,for some positive integer k.
For n=k+1,
= T1+T2+...+Tk+Tk+1
=(k+1)! -1 +(k+1)[(k+1)!]
=(k+1)![1+(k+1)] -1
=(k+1)! (k+2) -1
=(k+2)! -1
∴The statement is also true for n=k+1 if it is true for n=k.
By the principle of mathematical induction,the statement is true for all positive integer n.

T_1 = 1(1!) = 1 = 2! - 1

Assume \sum_{i = 1}^k = (k + 1)! - 1,
\sum_{i = 1}^{k + 1}
= (k + 1)! - 1 + T_{k + 1}
= (k + 1)! - 1 + (k + 1) * (k + 1)!
= ((k + 1) + 1) * (k + 1)! - 1

By the principle of mathematical induction, \sum_{i = 1}^{n} = (n + 1)! - 1 for all positive integers n



2008-08-06 13:24:48 補充：
我想問, 你o地之後o既答案同我呢個有咩分別呢?
唔好話我聽多o左o的"P(1) is true", "The statement is true for n=1"
而家係 [__教__] 人做功課唔係 [__幫人做__] 功課
求其一條書o既例子都有晒呢o的o野啦

2008-08-08 10:33:30 補充：
我冇自命清高
事實上, 你個答案o係我後面而最重要o既o野又同我o個個冇分別
你可以話我好cheap o地推銷我個答案添

而且我真係覺得你咁答法即係變相幫佢做功課囉