Prove the following by using the principle of mathematical induction for all
.![]()
Let P(n) : ![]()
I. For n = 1,
![]()
∴ P(1) is true
II. Let the statement be true for n = m, ![]()
∴ P(m):
...(i)
III. For n = m + 1,
![]()
or ![]()
or
... (ii)
From (i),
∴ ![]()
![]()
![]()
which is true
∴ P (m + 1) is true.
∴ P(m) is true
P(m + 1) is true.
Hence, by the principal of mathematical induction, P(n) is true for all ![]()



