F4 A.maths問題(2條,20分!)請清楚列出所有步驟!!!急...做好心

1. Let P(n) be the proposition that
1+2+3+.....+n=1/2(n)(n+1)

P(1): L.H.S. = 1
R.H.S. = 1/2(1)(1+1) = 1
Therefore, P(1) is true.

Assume P(k) is true for some integer k
i.e. 1+2+3+.....+k=1/2(k)(k+1)

P(k+1) : 1+2+3+.....+k+(k+1)
=1/2(k)(k+1)+(k+1)
=(k^2 + 3k + 2) / 2
= (k+1)(k+2) / (2)
Therefore, P(k+1) is true.

By Mathematics Induction,
P(n) is true for all positive integers n.

2a)
Let P(n) be the proposition that
2(2)+3(2^2)+4(2^3)+........+(n+1) (2^n)=n(2^(n+1))

P(1) : L.H.S. = 4
R.H.S. = 1(2^1+1) = 4
Therefore,P(1) is true

Assume P(k) is true for some integer k
i.e. 2(2)+3(2^2)+4(2^3)+...+(k+1) (2^k)=k(2^(k+1))

P(k+1) : 2(2)+3(2^2)+4(2^3)+...+(k+1) (2^k)+(k+2)(2^(k+1))
=k(2^(k+1)) )+(k+2)(2^(k+1))
=2^(k+1)(2k+2)
=(2^(k+2))(k+1)
Therefore, P(k+1) is true.

By Mathematics Induction,
P(n) is true for all positive integers n.

b) Put n=98 into (a)
2(2)+3(2^2)+4(2^3)+........+99 (2^98)=98(2^99)

1(2)+2(2^2)+3(2^3)+.....+98(2^98)
= 98(2^99) - [2(2^(98-1)) / (2-1)]
= 98(2^99) - (2^99 - 2)
=97(2^99) - 2


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1. Let S(n) be the statement 1+2+3+.....+n=n(n +1)/2
When n = 1, LHS = 1
RHS = 1(2)/2 = 1 = LHS
S(1) is true.
Assume S(k) is true, i.e. 1+2+3+.....+k=k(k+1)
When n = k+1, LHS = 1+2+3+.....+k + (k+1)
=k(k+1)/2 + k+1
= (k+1)(k/2 + 1)
=[(k+1)/2] (k+2)
=(k+1)(k+2)/2
S(k+1) is true.
By the principle of mathematical induction, the statement is true for all positive integers n.
2. (a) Let S(n) be the statement 2(2)+3(2^2)+4(2^3)+........+(n+1) (2^n)=n(2^(n+1))
When n = 1, LHS = (1+1)2^1 = 2(2) = 4
RHS = 1(2^2) = 4 = LHS
S(1) is true.
Assume S(k) is true, i.e. 2(2)+3(2^2)+4(2^3)+........+(k+1) (2^k)=k(2^(k+1))
When n = k+1, LHS = 2(2)+3(2^2)+4(2^3)+........+(k+1) (2^k) + (k+2)(2^(k+1))
= k(2^(k+1)) + (k+2)(2^(k+1))
= 2^(k+1) [k+(k+2)]
=2^(k+1) * 2(k+1)
= (k+1)* 2^(k+2)
S(k+1) is true.
By the principle of mathematical induction, the statement is true for all positive integers n.
(b) 1(2)+2(2^2)+3(2^3)+.....+98(2^98)
=[2(2)-2]+[3(2^2)-2^2]+[4(2^3)-2^3]+........+[(99) (2^98)-2^98]
=2(2)+3(2^2)+4(2^3)+........+99(2^98) - (2+2^2+2^3 + ...+2^98)
=98(2^99) - [2(2^98 - 1)/(2-1)]
=98(2^99) -(2^99 - 2)
=97(2^99) + 2