How do we find the sum of n terms of the series,
1+(4/7)+(7/49)+(10/343)+.....
I know that it can be written as a combination of an arithmetic and geometric progression.
1=(4+3r)/(7^m) ; (r=-1 , m=0)
4/7=(4+3r)/(7^m) ;(r=0,m=1)
But how do we find the sum of n terms?
S = 1 + 4 / 7 + 7 / 49 + 10 / 343...
S = 1 + 4 / 7 + 7 / 7² + 10 / 7³...
Divide both sides by 7
S / 7 = 1 / 7 + 4 / 7² + 7 / 7³ + 10 / 7...
S - S / 7 = 7 S / 7 - S / 7 =
6 S / 7 = 1 + 4 / 7 - 1 / 7 + 7 / 7² - 4 / 7² + 10 / 7³ - 7 / 7³...
6 S / 7 = 1 + 3 / 7 + 3 / 7² + 3 / 7³...
Divide both sides by 3
( 6 S / 7 ) / 3 = 1 / 3 + ( 3 / 7 ) / 3 + ( 3 / 7² ) / 3 + ( 3 / 7³ ) / 3...
6 S / 3 ∙ 7 = 1 / 3 + 3 / 3 ∙ 7 + 3 / 3 ∙ 7² + 3 / 3 ∙ 7³...
6 S / 21 = 1 / 3 + 1 / 7 + 1 / 7² + 1 / 7³...
6 S / 21 = 1 / 3 + ( 1 / 7 + 1 / 7² + 1 / 7³...)
The terms in the bracket form a infinite geometric progression.
a , a r , a r²...
In this case: a = 1 / 7, r = 1 / 7
a , a r , a r²...
1 / 7 , ( 1 / 7 ) ∙ ( 1 / 7 ) , ( 1 / 7 ) ∙ ( 1 / 7 )²...
1 / 7 , 1 / 7 ∙ 7 , 1 / ∙ 7²...
1 / 7 , 1 / 7² , 1 / 7³...
Now:
6 S / 21 = 1 / 3 + ( 1 / 7 + 1 / 7² + 1 / 7³...)
( 1 / 7 + 1 / 7² + 1 / 7³... = = S∞
so:
6 S / 21 = 1 / 3 + S∞
The sum of the infinite geometric progression:
S∞ = a / ( 1 - r )
In this case: a = 1 / 7, r = 1 / 7
S∞ = ( 1 / 7 ) / ( 1 - 1 / 7 ) =
( 1 / 7 ) / ( 7 / 7 - 1 / 7 ) =
( 1 / 7 ) / ( 6 / 7 ) =
1 ∙ 7 / 6 ∙ 7 = 1 / 6
S∞ = 1 / 6
Now:
6 S / 21 = 1 / 3 + S∞
6 S / 21 = 1 / 3 + 1 / 6
6 S / 21 = 2 / 6 + 1 / 6
6 S / 21 = 3 / 6
6 S / 21 = 3 ∙ 1 / 3 ∙ 2
6 S / 21 = 1 / 2
Multiply both sides by 21
6 S = 21 / 2
Divide both sides by 6
S = ( 21 / 2 ) / 6
S = 21 / 2 ∙ 6
S = 21 / 12
S = 3 ∙ 7 / 3 ∙ 4 = 7 / 4 = 1.75
What are you still have not
Ano biotic and abiotic factors
S = 1 + 4 / 7 + 7 / 49 + 10 / 343...
S = 1 + 4 / 7 + 7 / 7² + 10 / 7³...
Divide both sides by 7
S / 7 = 1 / 7 + 4 / 7² + 7 / 7³ + 10 / 7⁴...
My typo again.
Now:
6 S / 21 = 1 / 3 + ( 1 / 7 + 1 / 7² + 1 / 7³...)
( 1 / 7 + 1 / 7² + 1 / 7³... ) = S∞
As we are finding the sum of n terms,shouldn't the sum of the geometric progression should be,
a{ [(r)^(n)- 1]/[r-1] } , where a=(1/7) and r=(1/7) ?
You can't use the sum of n terms in this case.
1 / 7 , 1 / 7² , 1 / 7³...
is a infinite geometric progression:
1 / 7 + 1 / 7² + 1 / 7³ + 1 / 7⁴+ 1 / 7⁵ + 1 / 7⁶ + 1 / 7⁷ + 1 / 7⁸ + 1 / 7⁹...
Therefore you must use the sum of the infinite geometric progression.