Math

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?

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  1. 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

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  2. 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...

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  3. My typo again.

    Now:

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

    ( 1 / 7 + 1 / 7² + 1 / 7³... ) = S∞

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  4. 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) ?

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  5. 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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  6. What are you still have not

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  7. Ano biotic and abiotic factors

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