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



