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Math Help - Recurrence Relations

  1. #1
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    Recurrence Relations

    The formula 1+2+3+...+n = (n(n+1))/2 is true for all integers n>=1. Use this fact to solve each of the following problems:
    If n is an integer and n>=1, find a formula for the expression
    3+3(2)+3(3)+...+3(n)+n

    Is this correct?
    3+3((n(n+1))/2)+3
    3+(3/2)(n(n+1))+3
    3+(3/2)(n^2+n)+3
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  2. #2
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    Re: Recurrence Relations

    Quote Originally Posted by lovesmath View Post
    The formula 1+2+3+...+n = (n(n+1))/2 is true for all integers n>=1. Use this fact to solve each of the following problems: If n is an integer and n>=1, find a formula for the expression
    3+3(2)+3(3)+...+3(n)+n
    3(1)+3(2)+\cdots+3(n)+n=3\frac{n(n+1)}{2}+n=~?
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    Re: Recurrence Relations

    The answer to 3+2+4+6+8+...+2n (if n is an integer and n>=1) is n^2+n+3. I thought since the original question I posted began with a 3, it would also add 3 like this example. If not, why?
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    Re: Recurrence Relations

    Quote Originally Posted by lovesmath View Post
    The answer to 3+2+4+6+8+...+2n (if n is an integer and n>=1) is n^2+n+3. I thought since the original question I posted began with a 3, it would also add 3 like this example. If not, why?
    Look at what you did post.
    3+3(2)+2(3)+\cdots+3(n)+\color{red}n.

    That is not at all the same as 3+2+4+6+8+...+2n.

    If you don't post correctly, how can we help? Do we guess?
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    Re: Recurrence Relations

    I was using the 3 + ... + 2n as an example. Why is it 3 + 2((n(n+1))2) = n^2 + n + 3? I don't understand where the 3 came from other than the fact that it is the first term in the sequence. I used that as an example because my problem, 3 + ... + 3n + n, also starts with a 3.
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    Re: Recurrence Relations

    Quote Originally Posted by lovesmath View Post
    I was using the 3 + ... + 2n as an example. Why is it 3 + 2((n(n+1))2) = n^2 + n + 3? I don't understand where the 3 came from other than the fact that it is the first term in the sequence. I used that as an example because my problem, 3 + ... + 3n + n, also starts with a 3.
    The expression 3 + ... + 2n is so poorly written that no one can read it.

    If you mean 3 + 2+ 4+... + 2n then that is the same as
    3+2(1)+2(2)+\cdots+2(n)=
    3+2(1+2+\cdots+n)=
    3+2\left(\frac{n(n+1)}{2}\right)
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