Use the principle of mathematical induction to prove that $a + (a + d) + (a + 2 d) + . . . . + [a + (n - 1) d] = n / 2 [2 a + (n - 1) d]$

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Solution

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]
Hence prove !

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