[rb30999]

Posting for the first time on pg and a problem straightaway

Hall and knight chapter on geometric progression, exercise no. V.b. ques 21 and 23

21. If Sn denotes the sum of a GP whose first term is a and common ratio r, find the sum of S1, S2, S3......S(2n-1). (I hope it is understood here that S(2n-1) means sum of 2n-1 terms)

23. If r<1 and and positive and m is a positive integer, show that

(2m+1)r^m * (1-r) < (1-r)^(2m+1)


I would really appreciate it if anyone could plz give me a detailed solution with each step explained

[implex]

Originally Posted by rb30999

Posting for the first time on pg and a problem straightaway

Hall and knight chapter on geometric progression, exercise no. V.b. ques 21 and 23

21. If Sn denotes the sum of a GP whose first term is a and common ratio r, find the sum of S1, S2, S3......S(2n-1). (I hope it is understood here that S(2n-1) means sum of 2n-1 terms)

23. If r<1 and and positive and m is a positive integer, show that

(2m+1)r^m * (1-r) < (1-r)^(2m+1)


I would really appreciate it if anyone could plz give me a detailed solution with each step explained

we know that S(k)=k[2a+(k-1)r]/2

now we have to sume S(k) for k=1 to 2n-1
S= a.(2n-1).2n/2+ r[ 2n.(2n-1)(4n+1)/6 -2n.(2n-1)/2]/2
=n.(2n-1)[a -r/2] +r.n.(2n-1).(4n+1)/6

we can further solve it if the options so demand

[rb30999]

Originally Posted by implex

we know that S(k)=k[2a+(k-1)r]/2

now we have to sume S(k) for k=1 to 2n-1
S= a.(2n-1).2n/2+ r[ 2n.(2n-1)(4n+1)/6 -2n.(2n-1)/2]/2
=n.(2n-1)[a -r/2] +r.n.(2n-1).(4n+1)/6

we can further solve it if the options so demand

Thanks for the reply implex but u see that the question is for geometric progression and not arithmetic progression. the formula u have used is for Sn of an AP

[implex]

Originally Posted by rb30999

Thanks for the reply implex but u see that the question is for geometric progression and not arithmetic progression. the formula u have used is for Sn of an AP

oh sorry man ..
I thought it to be AP
anyways will do it again

[implex]

S1=a
S2=a+ar
S3=a+ar+ar.r and so on

so when we need ths sum S=S1+S2+.... S(2n-1)=(2n-1)a+(2n-2)ar+(2n-3)....

clealry S is the sum of AGP

S-Sr=(2n-1)a -ar-ar.r-....-a.r^2n=a[ 2n- {1-r^(2n+1)}/(1-r)]
S =a[ 2n- {1-r^(2n+1)}/(1-r)]/(1-r)

[Mahip]

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