Question

If the first term ,common ratio and the sum of first n terms of a G.P be a,r and Sn respectively, find the value of S1+S2+S3+...+Sn?

Answer 1

See below.

Explanation:

`a sum_(k=0)^(n-1)r^k = S_n` then

`S_n = a((r^n-1)/(r-1))` because

`sum_(k=0)^(n-1)r^k = (r^n-1)/(r-1)`

and

`sum_(k=1)^nS_k = a/(r-1)sum_(k=1)^n(r^k-1) = a((r^(n+1)-1)/(r-1)^2-n/(r-1))`

Answer 2

`S_1 + S_2 + ... + S_n = (an)/(1-r) - a(1-r^n)/(1-r)^2`

Explanation:

If we have a GP with:

First Term: `\ \ \ \ \ \ \ \ \ "= a"`
Common Ratio: `\ \ "= r"`
Number of terms: `"= n"`

Then the sum of the first `n` term is given by the standard GP formula:

`S_n = a(1-r^n)/(1-r)`

The sum we seek is given by:

`T_n = sum_(r=1)^n S_r`
`\ \ \ \ = S_1 + S_2 + ... + S_n`
`\ \ \ \ = a(1-r^1)/(1-r) + a(1-r^2)/(1-r) + a(1-r^3)/(1-r) + ... + a(1-r^n)/(1-r)`
`\ \ \ \ = a{((1-r^1) + (1-r^2) + (1-r^3) + ... + (1-r^n))/(1-r)}`

`\ \ \ \ = a{( overbrace((1+1+...+1))^(n " terms") +overbrace((-r^1-r^2-...-r^n))^(n " terms") ) /(1-r)}`

`\ \ \ \ = a{( (1+1+...+1) - (r^1+r^2+r^3+...+r^n) ) /(1-r)}`

`\ \ \ \ = a{( n - (r^1+r^2+r^3+...+r^n) ) /(1-r)}`

`\ \ \ \ = (an)/(1-r) - {a(r^1+r^2+r^3+...+r^n)}/(1-r)`

`\ \ \ \ = (an)/(1-r) - S_n/(1-r)`

`\ \ \ \ = (an)/(1-r) - {a(1-r^n)/(1-r)}/(1-r)`

`\ \ \ \ = (an)/(1-r) - a(1-r^n)/(1-r)^2`