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# find the sum of the following series:(a+b)+(a^2+ab+b^2)+(a^3+a^2b+ab^2+b^3)+........

9 years ago

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.

9 years ago

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,

9 years ago

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 9 years ago

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,

Got it !! 