Question

using principle of mathamatical induction prove that a+(a+d)+(a+2d)+...........+[a+(n-1)d] =
n/2[2a+(n-1)d]

Answer 1

Let n=1
Then
a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = a
On the other
hand,
n/2[2a+(n-1)d] = 1/2 * 2a = a
So these expresions
coincide.

Supose that we have proved the identity for all k<n+1,
so
in particular, for k=n we have that

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] =
n/2[2a+(n-1)d]

We have to prove the identity for k=n+1, that
is

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd] = (n+1)/2 *
[2a+nd]

Notice that by induction
a+(a+d)+(a+2d)+ ... +[a+(n-1)d] +
[a+nd] =
n/2[2a+(n-1)d] + [a+nd] =
na + n(n-1) d/2 + a + nd
=
(n+1) a + (n^2/2 - n/2 + n) d =
(n+1) a + (n^2/2 + n/2) d
=
(n+1) a + (n+1) n d /2 =
(n+1)/2 [ 2 a + n d ]