Question

If Sn denotes the sum of n terms of a G.P. whose first term is a , common ratio r, find the sum of \[S_{1},S_{3},S_{5},............,S_{2n-1}\]

Answer 1

\(S_1=a\)
\(S_2=a+ar\)
\(S_3=a+ar+ar^2\) and so on
you want a closed form for this?

Answer 2

yes

Answer 3

\[\sum_{k=0}^nar^k=a\sum_{k=0}^nr^k=a(\frac{r^{n+1}-1}{r-1})\] should work

Answer 4

the ans. is \[\frac{a}{r-1}[\frac{r(r^{2n}-1)}{r^{2}-1}-n]\]

Answer 5

let a=1, for sake of ease
\[ S_1 = 1\\
S_3 = 1 + r + r^2 \\
S_5 = 1 + r + r^2 + r^3 + r^4 \\
S_7 = 1 + r + r^2 + r^3 + r^4 + r^5 + r^6 \\
.........\\
S_{2n-1} = 1 + r + r^2 + r^3 + r^4 + r^5 + r^6+ ... + r^{2n-2}
\]
Adding up
We get
\[ n + (n-1)(r+r^2) + (n-2)(r^3 + r^4) + ... + r^{2n-2}\]

Answer 6

n(1 + r + r^2 + r^3 + .. + r^(2n-2)) - ( r + 2r^3 + 3r^5 + 4r^7 ... + (n-1)r^(2n-3))
- (r^2 + 2r^4 +3 r^6 + ... + (n-1) r^(2n - 2))

Answer 7

i guess this part is solved
n(1 + r + r^2 + r^3 + .. + r^(2n-2))

Answer 8

( r + 2r^3 + 3r^5 + 4r^7 ... + (n-2)r^(2n-3)) = r(1 + 2r^2 + 3r^6 + 4 r^8 + ... + (n-2) r^(2n - 4))

Answer 9

let r^2 = x
x + 2x^2 + 3x^3 + ... +nx^n to evaluate this type of sequence,
divide it by x, you get
x( 1 + 2x + 3x^2 + ... + nx^(n-1))
x d/dx (x + x^2 +x^3 + ... + x^n)
x d/dx ((x^n - 1)/(x-1))

Answer 10

i hope you get the general idea on how to do this ....

Answer 11

well ... the sum is
http://www.wolframalpha.com/input/?i=x+D [%28%28x^n+-+1%29%2F%28x-1%29%29%2Cx]&dataset=&equal=Submit