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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)+........

find the sum of the following series:




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

Grade:10

5 Answers

Swapnil Saxena
102 Points
9 years ago

The answer for the question is 

(a (a^n-1))\/(a-1)+(b (b^n-1))\/(b-1)+(a b)\/(b-a) ((b (b^(n-1)-1))\/(b-1)+(a (a^(n-1)-1))\/(a-1))

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


Ashwin Muralidharan IIT Madras
290 Points
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,

Ashwin (IIT Madras).

Sathya
35 Points
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 Wink

Ashwin Muralidharan IIT Madras
290 Points
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,

Ashwin (IIT Madras).

Sathya
35 Points
9 years ago

Got it !! Wink

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