By mathematcal induction, prove that the following statements are true for all positive integers n.
>>可以只寫L.H.S.同R.H.S.<<因為太長....
Ques: 3 + 3^2 + 3^3 + ... + 3^n = 3/2 ( 3^n - 1 )
Mathematical induction (proofs)
2008-08-18 5:00 am
回答 (2)
2008-08-18 5:28 am
✔ 最佳答案
Mathematical induction:圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths398.jpg?t=1218979699
2008-08-17 21:30:49 補充：
http://hk.knowledge.yahoo.com/question/question?qid=7007090904245
You may refer to my previous answer.
參考: My Maths Knowledge
2008-08-18 5:44 am
Let P(n) be the proposition that "3 + 3^2 + 3^3 + ... + 3^n = 3/2 ( 3^n - 1 )"
When n=1,
LHS = 3
RHS = 3/2(3-1) = 3
As LHS = RHS, P(n) is true.
Assume that P(k) is true for some positive integers k
i.e. 3 + 3^2 + 3^3 + ... + 3^k = 3/2 ( 3^k - 1 )
When n=k+1,
LHS = 3 + 3^2 + 3^3 + ... + 3^k + 3^(k+1)
= 3/2( 3^k - 1 ) + 3^(k+1)
= 3/2(3^k) + 3^(k+1) - 3/2
= 3/2* 3^k(1+2*3/3) - 3/2
= 3/2* 3^(k+1) - 3/2
= 3/2(3^(k+1) -1)
= RHS
So P(k+1) is true
By the principle of mathematical induction, P(n) is true for all positive
integers n
When n=1,
LHS = 3
RHS = 3/2(3-1) = 3
As LHS = RHS, P(n) is true.
Assume that P(k) is true for some positive integers k
i.e. 3 + 3^2 + 3^3 + ... + 3^k = 3/2 ( 3^k - 1 )
When n=k+1,
LHS = 3 + 3^2 + 3^3 + ... + 3^k + 3^(k+1)
= 3/2( 3^k - 1 ) + 3^(k+1)
= 3/2(3^k) + 3^(k+1) - 3/2
= 3/2* 3^k(1+2*3/3) - 3/2
= 3/2* 3^(k+1) - 3/2
= 3/2(3^(k+1) -1)
= RHS
So P(k+1) is true
By the principle of mathematical induction, P(n) is true for all positive
integers n
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