Prove the following by the principle of mathematical induction:

a + (a + d) + (a + 2d) + … + (a + (n– 1)d)

P(n): a + (a + d) + (a + 2d) + … + (a + (n– 1)d) =

For n = 1

a = [2a + (1 - 1)d]

a = a

Since, P(n) is true for n =1,

Let P(n) is true for n = k, so

a + (a + d) + (a + 2d) + … + (a + (k– 1)d) = - - - - - (1)

We have to show that,

a + (a + d) + (a + 2d) + … + (a + (k– 1)d) + (a + (k)d) = [2a + kd]

Now,

{a + (a + d) + (a + 2d) + … + (a + (k– 1)d)} + (a + kd)

= [2a + (k - 1)d] + (a + kd) using equation

=

=

=

=

= [2a + kd]

Therefore, P(n) is true for n = k + 1

Hence, P(n) is true all n∈ N by PMI