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