F.4 A.MATH Mathematical Induction

Prove.by mathematical innduction,that
1x2+2x3+3x4+.......+n(n+1) = n(n+1)(n+2)/3
for all positive integers n.
當 n = 1
LHS = 1x2 = 2
RHS = n(n+1)(n+2)/3 = 1(1+1)(1+2)/3 = 2
當n=1時成立
設n = k時成立
則1x2+2x3+3x4+.......=k(k+1)(k+2)/3
當 n = k+1
LHS=1x2+2x3+3x4+......+k(k+1)+(k+1)(k+2)
= k(k+1)(k+2)/3 + (k+1)(k+2)
= k(k+1)(k+2)/3 + 3(k+1)(k+2)/3
= (k+1)(k+2)(k+3)/3
等於RHS所以對任何整數 n 均成立

Hence evaluate
1x3+2x4+3x5+........ +50x52.
1x(2+1)+2x(3+1)+3x(4+1)+........ +50x(51+1)
=1x2 + 2x3+3x4+…..50x51 + 1+2+3+…..50
所以等於
50(50+1)(50+2)/3 + (1+50)x50/2
= 45475

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

When n=1
LHS = 2
RHS = 1/3 (1)(1+1)(1+2) = 2

Assume that P(k) is true
ie, P(n) be 1x2+2x3+3x4+........+ k(k+1) = 1/3k(k+1)(k+2)
When n = k+1
1x2+2x3+3x4+........+ k(k+1) + (k+1)[(k+1)+1]
= 1/3k(k+1)(k+2) + (k+1)(k+2)
= 1/3(k+1) [k(k+2) + 3(k+2)]
= 1/3(k+1) [k^2+2k+3k+6]
= 1/3(k+1)(k+2)(k+3)
= 1/3(k+1) [(k+1)+1] [(k+1)+2]

Therefore, P(k+1) is true

By the principle of mathematic induction, P(n) is true for all positive integers n.

1x3+2x4+3x5+........+50x52
= (1x2+1)+(2x3+2)+(3x4+3)+......+(50x51+50)
= (1x2+2x3+3x4+......+50x51)+(1+2+3+......+50)
= 1/3 (50)(50+1)(50+2) + 1/2 (50)(50+1)
= 45475