# Answer on Real Analysis Question for Nevetha

Question #13983

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

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

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

Expert's answer

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 ]

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 ]

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