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    What's the difference between the result

    \displaystyle \sum_{r=1}^n r = \dfrac{n}{2} (n+1)

    and the result

    \displaystyle \sum_{r=1}^n b = nb
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    (Original post by huiop)
    What's the difference between the result

    \displaystyle \sum_{r=1}^n r = \dfrac{n}{2} (n+1)
    and the result

    \displaystyle \sum_{r=1}^n b = nb
    The first one is  1+2+3+...+n while the first one  \underbrace{b+b+b+b+...+b}_{n} .
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    (Original post by B_9710)
    The first one is  1+2+3+...+n while the first one  b+b+b+b+...+b where there are n b's.
    so how would i know which one to use in a question which says

    show that

    \displaystyle \sum_{r=1}^n (2r-1)^2 = \dfrac{n}{3} (4n^2 -1)

    so far i got to this \displaystyle 4\sum_{r=1}^n r^2 -4\sum_{r=1}^n r +\sum_{r=1}^n 1

    then i went to this \dfrac{4\times n}{6} (2n+1)(n+1)-4n+1

    so that 4n is wrong so how would i know to use standard result for r instead of b?
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    (Original post by huiop)
    so how would i know which one to use in a question which says

    show that

    \displaystyle \sum_{r=1}^n (2r-1)^2 = \dfrac{n}{3} (4n^2 -1)

    so far i got to this \displaystyle 4\sum_{r=1}^n r^2 -4\sum_{r=1}^n r +\sum_{r=1}^n 1

    then i went to this \dfrac{4xn}{6} (2n+1)(n+1)-4n+1

    so that 4n is wrong so how would i know to use standard result for r instead of b?
    Where does the x come from?

    \displaystyle 4\sum_{r=1}^n r^2 =4[\frac{1}{6}n(2n+1)(n+1)]
    \displaystyle -4\sum_{r=1}^n r =-4[\frac{1}{2}n(n+1)]
    \displaystyle \sum_{r=1}^n 1 =n
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    (Original post by RDKGames)
    Where does the x come from?

    \displaystyle 4\sum_{r=1}^n r^2 =4[\frac{1}{6}n(2n+1)(n+1)]
    \displaystyle -4\sum_{r=1}^n r =-4[\frac{1}{2}n(n+1)]
    \displaystyle \sum_{r=1}^n 1 =n
    whoops typing error i thought i could use x as a substitute for \times
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    (Original post by huiop)
    whoops typing error i thought i could use x as a substitute for \times
    As for the difference between r and b;

    If it's r, then you are adding 1+2+3+4+5+...+n because you are summing up the integers from r=1 up to r=n.

    If it's a constant (like b), then you are adding b+b+b+b+b+...+n because you are summing up a constant b, in a series of b, b, b, b, b, b, .... where every single term is b. So r=1 will be b, and r=2 will be b, and r=3 and so on up to r=n. Hence where nb comes from. As you are summing b an n amount of times.
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    (Original post by RDKGames)
    As for the difference between r and b;

    If it's r, then you are adding 1+2+3+4+5+...+n because you are summing up the integers from r=1 up to r=n.

    If it's a constant (like b), then you are adding b+b+b+b+b+...+n because you are summing up a constant b, in a series of b, b, b, b, b, b, .... where every single term is b. So r=1 will be be, and r=2 will be b, and r=3 and so on up to r=n. Hence where nb comes from. As you are summing b an n amount of times.
    ok thanks for clarification
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    (Original post by huiop)
    so how would i know which one to use in a question which says

    show that

    \displaystyle \sum_{r=1}^n (2r-1)^2 = \dfrac{n}{3} (4n^2 -1)

    so far i got to this \displaystyle 4\sum_{r=1}^n r^2 -4\sum_{r=1}^n r +\sum_{r=1}^n 1

    then i went to this \dfrac{4\times n}{6} (2n+1)(n+1)-4n+1

    so that 4n is wrong so how would i know to use standard result for r instead of b?
    Think about what this notation means. If you have  \displaystyle \sum_{r=1}^n 1 = 1+1+1+...+1 =n since there are n 1's. Using this notation, the sum is with respect to  r only. So if we have  \displaystle \sum_{r=1}^n b^2 = nb^2 not  (1/6)n(n+1)(2n+1) .
 
 
 
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