數學2001pp

1.
Two lines L1: x + y - 5=0 and L2: 2x - 3y = 0 intersect at a point A.Find the equations of the two lines passing through A whose distances from the origin are equal to 2.

Let (x + y - 5) + k(2x - 3y) = 0 be the required line.
(1 + 2k)x + (1 - 3k)y - 5 = 0

Distance between the required line and (0, 0) :
|(1 + 2k)(0) + (1 - 3k)(0) - 5| / √[(1 + 2k)² + (1 - 3k)²] = 2
|5| / √(1 + 4k + 4k² + 1 - 6k + 9k²) = 2
|5| / √(13k² - 2k + 2) = 2
[|5| / √(13k² - 2k + 2)]² = 2²
25/(13k² - 2k + 2) = 4
52k² - 8k + 8 = 25
52k² - 8k - 17 = 0
(26k - 17)(2k + 1) = 0
k = 17/26 or k = -1/2

The required lines are :
[1 + 2(17/26)]x + [1 - 3(17/26)]y - 5 = 0 or [1 + 2(-1/2)]x + [1 - 3(-1/2)]y -5 = 0
(30/13)x - (25/26)y - 5 = 0 or (5/2)y - 5 = 0
60x- 25y - 130 = 0 or y - 2 = 0


2.
Let P(n): 1x2 + 2x3 + 3x4 +......+ n(n+1)= (1/3)n(n+1)(n+2)

When n = 1:
L.S. = 1x2 = 2
R.S. = (1/3)(1)(1+1)(1+2) = 2
Since L.S. + R.S., P(1) is true.

Assume that P(k) is true, i.e.
1x2 + 2x3 + 3x4 +......+ k(k+1) = (1/3)k(k+1)(k+2)
To Prove that P(k+1) is also true, i.e.
1x2 + 2x3 + 3x4 +......+ (k+1)(k+2) = (1/3)(k+1)(k+2)(k+3)

Proof:
L.S.
= [1x2 + 2x3 + 3x4 +......+ k(k+1)] + (k+1)(k+2)
= (1/3)k(k+1)(k+2) + (k+1)(k+2)
= (k+1)(k+2)[(1/3)k + k]
= (k+1)(k+2)[(1/3)k + (1/3)3k]
= (1/3)(k+1)(k+2)(k+3)
= R.S.

Bythe principle of mathematical induction,
P(n) is true for all positive integers n.

1x3 + 2x4 + 3x5 +......+ 50x52
=1x(2+1) + 2x(3+1) + 3x(4+1) + ...... + 50x(51+1)
= (1x2 + 2x3 + 3x4 + ...... +50x51) + (1 + 2 + 3 + ...... + 50)
= (1/3)(50)(50+1)(50+2) + 50(50+1)/2
= 44200 + 1275
= 45475