Asked by Ben

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?
Answered by Bosnian
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







Answered by Bosnian
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...
Answered by Bosnian
My typo again.


Now:

6 S / 21 = 1 / 3 + ( 1 / 7 + 1 / 7² + 1 / 7³...)

( 1 / 7 + 1 / 7² + 1 / 7³... ) = S∞
Answered by Ben
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) ?
Answered by Bosnian
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.

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