Question

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

Answer 1

Hope this helps you.....

Answer 2

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

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]  

Hence prove !