Prove by mathematical induction?

Prove by mathematical induction:

3+7+11+...(4n-1)=n(2n+1)

Please show me steps and the answer. I don't really get this...

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  • Puggy
    Lv 7
    10 years ago
    Best Answer

    3 + 7 + 11 + ... + (4n - 1) = n(2n + 1)

    Proof (by induction)

    Base Case: Let n = 1. Then

    LHS = 3 ... (4(1) - 1)

    = 3 ... 3

    = 3

    RHS = (1)(2*1 + 1)

    = (1)(2 + 1)

    = (1)(3)

    = 3

    LHS = RHS, so the formula holds true for n = 1.

    Induction Hypothesis: Assume the formula holds true for up to n = k. That is, assume that

    3 + 7 + 11 + ... + (4k - 1) = k(2k + 1) for some value k.

    (We want to prove that the formula holds true for n = k + 1, i.e.

    3 + 7 + 11 + ... (4(k + 1) - 1) = (k + 1)(2(k + 1) + 1) )

    BUT, what ARE the sum of the (k + 1) terms equal to? Answer: They are equal to the sum of the k terms, plus the (k + 1) term. To put algebraically,

    3 + 7 + 11 + ... (4(k + 1) - 1) = [ 3 + 7 + 11 + ... + (4k - 1) ] + ( 4(k + 1) - 1 )

    See what I did there? It's like saying 1 + 2 + 3 + 4 + ... 10 = [1 + 2 + 3 + 4 + ... + 9] + 10.

    I'm just explicitly stating the second last term.

    Going back to that step, the expression in square brackets is precisely our induction hypothesis. Since we are assuming that the formula holds true for up to n = k, we are going to replace the square bracketed summation with k(2k + 1).

    3 + 7 + 11 + ... (4(k + 1) - 1) = [ k(2k + 1) ] + ( 4(k + 1) - 1 )

    From here, simplify the right hand side.

    = (2k^2 + k) + 4k + 4 - 1

    = 2k^2 + k + 4k + 3

    = 2k^2 + 5k + 3

    = (k + 1)(2k + 3)

    = (k + 1)(2(k + 1) + 1)

    Showing that

    3 + 7 + 11 + ... (4(k + 1) - 1) = (k + 1)(2(k + 1) + 1)

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

    Therefore, by assuming that the formula holds true for up to n = k, and that we were able to derive that the formula holds true for n = k + 1, by the Principle of Mathematical Induction,

    3 + 7 + 11 + ... + (4n - 1) = n(2n + 1)

    -----------------------------------------------

    The concept of mathematical induction. By assuming that the formula holds true for up to n = k, and then derivative that it holds true for n = k + 1, that means that

    if the formula holds true for n = 6, then it holds true for n = 6 + 1 = 7.

    It holds true for 7? It also holds true for 8!

    Holds true for 8? Also 9!

    9? 10 also!

    10? 11!

    That's precisely why, if we assume it is true for n = k, and then prove that it holds true for n = k + 1, that the formula holds true for all natural numbers.

  • 10 years ago

    any induction question.we prove for n = 1.assume the statement to be true for n = k.then we try and prove that for n = k+1 the statement is true.since it is true for n= 1 it must be true for n = 1+1 =2 and so on

    P ( n) be the statement 3 + 7+11+......( 4n-1) = n ( 2n + 1)

    p(1) is true ie for n = 1 LHS = 3 and RHS = 1 ( 2+ 1) = 3 LHS and RHS are equal

    let P(k) be true ie 3+ 7+11.....+ (4k-1) = k ( 2k+1)

    for n = k+1 we have to show that P(k+1) = 3 + 7+ 11+ ..........+ (4k-1) + [4(k+1)-1] = ( k+ 1) (2 k + 3)

    LHS = k (2k+ 1) + 4k+ 3

    =2k^2 + 5k + 3 = (k +1) ( 2k + 3 )= RHS

    hence by induction the statement is true for all values of n

  • 10 years ago

    Let P(n) be the proposition

    Base Step: P(1)= 1*(2*1+1)=3 is true. Base step is completed.

    Induction Step: Assume P(k) is true, we must show that P(k+1) must also be true to prove this formula.

    P(k) is 3+7+11+...+(4k-1)=k(2k+1)

    add 4(k+1)-1 to both sides

    Now right hand side is equal to 2k^2+5k+3 which is equal to (k+1)(2k+3) which is equal to P(k+1).

  • Pere D
    Lv 5
    10 years ago

    The pattern has the addends increasing by 4, so each pairing 1st plus last, 2nd plus 2nd from last, et cetera will equal 4n + 2. There are n/2 such pairings, so the grand total will = (n/2)(4n + 2). Distributing the division, this = n(2n + 1).

    [The division by two accommodates odd and even numbers of addends.]

    Edit- When pattern hacking, one notices that for odd numbers of addends the median is the mean. If n is odd, the median's place is ordinally (n + 1)/2 and has value, as mean, of 4((n + 1)/2) - 1 = 2n + 1. Pairs of addends equidistant from the median will have as their average 2n + 1, so the total of n addends of such average value is n(2n +1).

    [For even numbers of addends, the mean falls (n/2)th and (n/2 + 1)th addends but is still ((4n/2 - 1) + (4n/2 + 4 - 1))/2 = 2n + 1 and in good standing for n addends.]

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  • Ken
    Lv 5
    10 years ago

    let us suppose that it works for the sequence up to (4k-1) and =k(2k+1)

    now if we add the next term in the sequence to both sides

    3+7+11+....+(4k-1)+(4(k+1)-1)=k(2k+1)+4k+3

    the rhs =2k^2+5k+3

    which =(k+1)[2k+3]

    which is exactly what we would have if we substitute k+1 for k in the original expression.

    So if it works for k then it works for k+1 ,providing k is a positive integer

    and we know it works for k=3 since

    3+7+11=3(7)=21

    and since we can let k=n that should do it.

  • 10 years ago

    S_(k + 1) = S_k + t_(k+ 1)

    S_(k + 1) = k(2k + 1) + (4k + 3)

    S_(k + 1) = 2k^2 + 5k + 3

    S_(k + 1) = 2k^2 + 2k + 3k + 3

    S_(k + 1) = 2k(k + 1) + 3(k + 1)

    S_(k + 1) = (2k + 3)(k + 1)

    Source(s): my brain
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