唔識計amaths~

Let P(n) be n^3+5n is divisible by 6 for all natural numbers n .

When n = 1,

1^3+5(1)
=1+5
=6
which is divisible by 6
So P(0) is true.

Assume P(k) is true, i.e.k^3+5(k) is divisible, i.e.
k^3+5(k) = 3m where m is a natural number and k is any natural number.

When n = k+1,
(k+1)^3+5(k+1)
= (k^3+3k²+3k+1) + 5k + 5
= (k^3 + 5k) + 3k² + 3k + 1
= 6m + 3k² + 3k + 6 .............. by assumsion of P(k)
= 6m + 3(k)(k+1) + 6 .... (*)

As for k(k+1) either k or k+1 is even number,
so k(k+1) is divisible by 2,
so k(k+1) = 2T, where T = k(k+1)/2

(* continued)
= 6m + 6T + 6
= 6(m+T+1)
which is divisible by 6
So P(n+1) is true.

By Mathematical Induction, n^3+5n is divisible by 6 for all natural numbers n.
2
let the root is a
then
a^2+ka-6=0...(1)
2(1/a)^2+k(1/a)-1=0...(2)
from (2)
2+ka-a^2=0...(3)
from (1)
a^2=6-ka
sub into (3)
2+ka-6+ka=0
2ka-4=0
ka=2
k=2/a
sub into (1)
a^2=6-ka
a^2=6-(2/a)a
a^2=4
a=2 or a=-2
k=1 or -1

by mathematical induction
n^3+5n is divisible by6 for all natural numbers n

[Solution]

Let P(n) be the proposition "n^3+5n is divisible by 6".
For n=1, 1^3+5(1) = 6 = (6)(1) , which is divisible by 6.
∴P(1) is true.

Assume that P(k) is true,
i.e. k^3+5k = 6M, M is an integer.

For n=k+1,
(k+1)^3+5(k+1)
=k^3+3k^2+3k+1+5k+5
=k^3+5k+3k^2+3k+6
=6M+6+3(k^2+k)
=6(M+1)+3k(k+1)
=6(M+1)+3(2R) [where R is an integer, ∵k(k+1) must be divisible by 2]
=6(M+1+2R), which is also divisible by 6

∴P(k+1) is also true.

By the first principle of mathematical induction, P(n) is true for all natural numbers n.


另一條
given that one root of the equation x^2+kx-6=0 is the reciprocal of a root of the equation 2x^2+kx-1=0,find the values of k

[Solution]

Let q be the one root of the equation x^2+kx-6=0, so the root of another equation 2x^2+kx-1=0 is 1/q.

Put q and 1/q into the corresponding equations, i.e.
q^2+kq-6=0 ........................................(1)

2(1/q)^2+k(1/q)-1=0
2/(q^2)+k/q-1=0
2+kq-q^2=0 [∵q≠0] ............................(2)

(1)-(2):
2q^2-8=0
q^2=4
q=2 or q=-2

Put q=2 into (1).
2^2+k(2)-6=0
2k=2
k=1

Put q=-2 into (1).
(-2)^2+k(-2)-6=0
2k=-2
k=-1

Overall solution: k=1 or k=-1

For n=1
1^3+5(1)=6
The statement is true for n=1

Assume k^3+5k=6M for some +ve integer k and M is integer
Then
(k+1)^3+5(k+1)
=k^3+3k^3+3k+1+5k+5
=(k^3+5k)+(3k^2+3k+6)
=6M+3k(k+1)+6
=6M+6N+6
(Note that k(k+1) is even for all +ve integers k, let k(k+1)=2N)
The statement is true for n=k+1

By principle of induction, the statement is true for all +ve integers n



Let λ be a real number such that
λ is a root of x^2+kx-6=0 and 1/λ is a root of 2x^2+kx-1=0

λ^2+kλ-6=0 ------(1)
2(1/λ)^2+k(1/λ)-1=0
i.e. λ^2-kλ-2=0 ---(2)

(1)+(2),
2λ^2-8=0
λ=±2

k
=(6-λ^2)/λ
=(6-4)/±2
=±1