Question

if |a|,|b|<1, then find the sum of the series:-
a(a+b) + a^2(a^2+b^2) +a^3(a^3+b^3) + ........infinity.

Note : this is not in GP series.
Then how can we find the sum of series

Answer 1

@ExperimentX

Answer 2

plse help me

Answer 3

@Shubhamsrg
Plse help me

Answer 4

@ash2326

Answer 5

plse help me

Answer 6

well this is not in GP series but you can convert it into 2 separate GPs

why dont you expand all the brackets ?

Answer 7

how

Answer 8

a(a+b) = a^2 + ab
a^2 (a^2 + b^2 ) = a^4 + a^2 b^2
like that.

Answer 9

following?

Answer 10

wait ..i m trying

Answer 11

yes ..i m getting sam e vat u have written , vat after that

Answer 12

Why didnt you simply write it here? What was the need for a doc file ? o.O

Answer 13

anyways, try forming 2 separate GPs
1 would be
a^2+ a^4 + a^6 ....
can you tell me second ?

Answer 14

ab+a^2b^2+a^3b^3

Answer 15

but we will take common ab from second

Answer 16

ab(1+ab+a^2b^2)

Answer 17

you may, though there is no need..

Answer 18

k

Answer 19

so now you have 2 GPs, both infinite and common ratio < 1 ..so you can do further right ?

Answer 20

yes..

Answer 21

thank u

Answer 22

on e thing more..it means we will be getting two G.P. + two answers.

Answer 23

i m getting answer, while solving first G.P
a^2/(1-a^2)
and after solving second one , i get
ab/(1-ab)

Answer 24

so final answer will be sum of both those sums, ofcorse.

Answer 25

[a^2/(1-a^2)]+[ab/(1-ab)]

Answer 26

yep.

Answer 27

thanks again