find the sum of the following series:

(a+b)+(a^2+ab+b^2)+(a^3+a^2b+ab^2+b^3)+........

The answer for the question is 

(For all a,b elonging to real no except 1 ) where n is the no of terms in the series.

Hi Satyaram,

Just multiply and divide the expression by (a-b) and sum it.

The expression will become

S = [(a²-b²) + (a³-b³) + (a⁴-b⁴) +..........]/(a-b)

So S = [ (a²+a³+....) - (b²+b³+.....) ] / (a-b)

So S = [ a²/(1-a) - b²/(1-b) ] / (a-b)

or S = (a+b-ab)/[(1-a)(1-b)]

Hope it helps.

Regards,

Ashwin (IIT Madras).

How ,

So S = [ (a2+a3+....) - (b2+b3+.....) ] / (a-b)

Became,

So S = [ a2/(1-a) - b2/(1-b) ] / (a-b) ?

Pls explain...And thnks for ur answer 

Dude,

Sum of infinite GP

1+a+a²+a³+...... = 1/(1-a) ----------- (Remember the Geometric Progressions formula)

So a²+a³+a⁴+.... = a²/(1-a).

Regards,

Ashwin (IIT Madras).

Got it !!