數學證明題：設 P、Q、R 分別是等差數列的第p項...

Question

設 P、Q、R 分別是等差數列的第p項，第q項，第r項，證明： P(q-r) + Q(r-p) + R(p-q) = 0

Answer 1

假設
P = a1 + (p - 1)*d
Q = a1 + (q - 1)*d
R = a1 + (r - 1)*d

P(q-r) + Q(r-p) + R(p-q)
= p*(R - Q) + q*(P - R) + r*(Q - P)
= p*[ (r - q)*d] + q*[ (p - r)*d ] + r*[ (q - p)*d ]
= prd - pqd + pqd - qrd + rqd - prd
= 0

Answer 2

令公差為 d
P(q-r) + Q(r-p) + R(p-q)
=P(q-r) + [P+(q-p)d](r-p) + [P+(r-p)d](p-q)
=P(q-r + r-p + p-q) + d(r-p)[(q-p)+(p-q)]
= 0