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deep01

Find the sum upto the infinite terms of the series

1+(1+a)x+(1+a+a²)x²+(1+a+a²+a³)x³+-----------

pls explain ur answer

vish0001

hey this can be done as follows

the given series is summation of the followinf series from 0 to n

(1 + a + arsquare ................ ar raised to power k-1) x to power k-1
find the sum of the the abover seriees first and then summate it over the whole range using possibly properties of GP
hope u understood 1

deep01

but we have to find the sum upto infinite terms not upto k terms??

vish0001

yaar, but there is a formula for infinte terms as well ! use that ! u will get it !

vish0001

did u get it ?? yaar , posting the whole sol is a little troublesome ! hope u understood ! please ask doubt if any !

ankur_gupta_ntse_scholar

general term= the nth term of serie, n belong to natural number..

({a^n -1)/(a-1)}x^(n-1)

break it up :

a^n*x^(n-1)/a-1 - x^(n-1)/(a-1)/-1

well the first one is a GP with common ratio ax and the second with common ratio x..

using formulae for infinite terms we have - sum=a/1-r

a=first term
r=common ratio

so the sum is

1/(a-1){a/ax-1 - 1/x-1 }

well i hope the answer is right...do tell me if it is wrong..And if correct plzzz rate me..

Ankur

truly

ques.....1+(1+a)x+(1+a+a²)x²+(1+a+a²+a³)x³+-----------

breaking up

(1+x+x²+x³......) + a (x+x²+x³+x⁴ .....) + a² (x²+x³+x⁴+x⁵......) + ..........

=> (1+x+x²+x³......) + ax (1+x+x²+x³ .....) + a²x² (1+x+x²+x³.....) + ..........

=> (1+x+x²+x³......).(1 + ax + a²x² + a³x³+ ..........)

=> [1/(1-x)].[1/(1-ax)]

=>1/(1-x)(1-ax)

PARSHAD

taking the given series as
S =1+ (1+a)x+(1+a+a^2)x^2+(1+a+a^2+a^3)x^3+-----------
-S*x = -x(1) - x((1+a)x)+ - x((1+a+a2)x^2) -x((1+a+a2+a3)x^3)+----
----------------------------------------------------------------------------------------------------------------------
S(1-x) =1 + ax + a^2x^2 +a^3x^3......................

now this is an GP so sum upto infinity of an GP is 1/(1-r) where r is common ratio
here the common ratio is ax
so

S(1-x) =1/(1-ax)
so S= 1/(1-ax)(1-x)

rajat

S = 1 + (1+a)x + (1+a+a^2)x^2 + (1+a+a^2+a^3)x^3..............1

xS = x + (1+a)x^2 + (1+a+a^2)x^3 + ...................................2

subtract 2 from 1

S(1-x) = 1 + ax + (ax)^2 + (ax)^3 ..............................

= > S(1-x) = 1/1-ax ...........................[ summation of infinite GP with ratio =ax]

=> S = 1/(1-ax)(1-x)

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