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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)+........
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 ,
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).
Got it !!
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