1+r+r^2+r^3+ ... +r^n=[1-r^(n+1)]/(1-r)?

Prove geometric series formula:

For any real number r not equal to 0 and all integers n greater than or equal to 0,

1+r+r^2+r^3+ ... +r^n=[1-r^(n+1)]/(1-r)

(Here r^n denotes the n'th power of n.)

For any real number r not equal to 0 and all integers n greater than or equal to 0,

1+r+r^2+r^3+ ... +r^n=[1-r^(n+1)]/(1-r)

(Here r^n denotes the n'th power of n.)

4 Answers

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  • Puggy
    Lv 7
    1 decade ago
    Favorite Answer

    1 + r + r^2 + r^3 + ... + r^n = [1 - r^(n + 1)] / (1 - r), for n >= 0.

    What we want to do is solve this using mathematical induction, start at n = 0.

    Base case:

    Let n = 0. Then, the left hand side will be

    LHS = 1... 1 = 1, and

    RHS = [1 - r^(0 + 1)] / (1 - r) = [1 - r] / [1 - r] = 1

    So the formula holds true for n = 0

    Inductive step:

    Assume the formula holds true for n = k. That is,

    1 + r + r^2 + r^3 + ... + r^k = [1 - r^(k + 1)] / (1 - r)

    Then the sum of the first (k + 1) terms would be the same as the sum of the first k terms, plus the (k + 1)th term. That is

    1 + r + r^2 + r^3 + ... + r^(k + 1) = {1 + r + r^2 + ... r^k} + r^(k + 1)

    But, look at the terms I placed in the { } brackets; by our induction hypothesis, this is equal to [1 - r^(k + 1)] / (1 - r). Therefore

    1 + r + r^2 + r^3 + ... + r^(k + 1) = [1 - r^(k + 1)] / (1 - r) + r^(k + 1)

    Putting them under a common denominator,

    1 + r + r^2 + r^3 + ... + r^(k + 1) =

    [1 - r^(k + 1)] / (1 - r) + (1 - r) r^(k + 1) / (1 - r)

    Putting them under a single denominator,

    1 + r + r^2 + r^3 + ... + r^(k + 1) =

    [1 - r^(k + 1) + (1 - r) r^(k + 1)] / (1 - r)

    Distributing the r^(k + 1) over the (1 - r), we have

    1 + r + r^2 + r^3 + ... + r^(k + 1) =

    [1 - r^(k + 1) + (r^(k + 1) - r*r^(k + 1)] / (1 - r)

    Simplifying some more, note that r*r(k + 1) would be r^(k + 2).

    1 + r + r^2 + r^3 + ... + r^(k + 1) =

    [1 - r^(k + 1) + r^(k + 1) - r^(k + 2) ] / (1 - r)

    But, we have two terms that cancel each other out.

    1 + r + r^2 + r^3 + ... + r^(k + 1) = [1 - r^(k + 2)] / (1 - r)

    OR,

    1 + r + r^2 + r^3 + ... + r^(k + 1) = [1 - r^( (k + 1) + 1 ) / (1 - r)

    Which proves that the formula holds true for n = k + 1.

    Therefore, by the principal of mathematical induction, the formula holds true for all n > = 0.

  • ?
    Lv 4
    4 years ago

    words=0:a million:1000 %change the 1000 as had to work out what the sum procedures r= %Set inspite of r is an same as right here sum=0; for i = a million to length(words) sum=sum+a million / r^words(i); end sum %this may output your answer to the command window. Edit: After filing this the day previous to this, i realized it may likely be extra effective to have the loop run till the quantity added in a unmarried generation became less than a particular quantity.

  • Philo
    Lv 7
    1 decade ago

    You don't have to ask twice.

    Let S = 1 + r + r² + ... + r^n

    multiply by r, get Sr = r + r² + r^3 + ... + r^n + r^(n+1)

    subtract the 2 equations, lining up like terms.

    S - Sr = 1 + r - r + r² - r² + ... + r^n - r^n - r^(n+1)

    S - Sr = 1 - r^(n+1)

    S(1 - r) = 1 - r^(n+1)

    S = [1 - r^(n+1)] / (1 - r)

  • Anonymous
    1 decade ago

    -34.759

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