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Thread: Number pattern

  1. #1
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    Number pattern

    Hi, The question I have is:

    Express a single generality that generalises all three related observations that
    $\displaystyle 1+3+5+3+1=3^2+2^2$
    $\displaystyle 1+3+5+7+9+7+5+3+1=5^2+4^2$
    $\displaystyle 1+3+1=2^2+1^2$
    and show that your generalisation always holds true.

    I know that the answer (i.e. RHS of =) is always $\displaystyle \frac{(no. of different terms)^2}{2}+(\frac{(no. of different terms)}{2}-1)^2$ (supposed to say no. of different terms)

    I also know that LHS is $\displaystyle a+(a+2)+(a+4)+...+(a+4)+(a+2)+a$, where a=1


    but I'm not sure how to show all of this generally, so that it works for every case . Any ideas?
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  2. #2
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    Have you studies arithmetic series before ? If you have you should without much difficulty be able to write down the formula for the sum of the first n odd numbers.

    do you see how this relates to your problem ?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    1+3+5+...+(2k-1)+(2k+1)+(2k-1)+...+5+3+2+1

    1+3+5+...+(2k-1)=(2k-1)+...+5+3+2+1=(1+{2k-1}}*k/2=k^2

    Therefor:

    1+3+5+...+(2k-1)+(2k+1)+(2k-1)+...+5+3+2+1=

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

    ={k^2 + (2k+1) } +k^2 =

    =(k+1)^2+k^2
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  4. #4
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    Hello, cozza!

    bobak is absolutely correct . . . Did you catch his hint?


    Express a single generality that generalises all three related observations that:

    . . $\displaystyle \begin{array}{ccc}1 + 3 + 1 &=& 2^2 + 1^2 \\
    1+3+5+3+1 &=& 3^2+2^2 \\
    1+3+5+7+9+7+5+3+1 &=& 5^2+4^2 \end{array}$

    and show that your generalisation always holds true.

    Look at the left side:

    . . $\displaystyle \underbrace{1 + 3 + 5 + 7 + 9}_{\text{sum of first 5 odd numbers}} + \underbrace{7 + 5 + 3 + 1}_{\text{sum of first 4 odd numbers}} $



    In general, we have: .$\displaystyle \underbrace{1 + 3 + 5 + 7 + \hdots + (2k-1)}_{\text{sum of first }k\text{ odd numbers}}$

    This is a arithmetic series with:
    . . first term $\displaystyle a = 1$, common difference $\displaystyle d = 2$, and $\displaystyle n = k$ terms.

    Its sum is: .$\displaystyle S \:=\:\dfrac{k}{2}\bigg[2(1) + (k-1)2\bigg] \;=\;k^2$

    Hence, the sum of the first $\displaystyle k$ odd numbers is $\displaystyle k^2.$ .[1]



    Therefore, we have:

    . . $\displaystyle \underbrace{1 + 3 + 5 + \hdots + (2n-1)}_{\text{sum of first }n\text{ odd numbers}} + \underbrace{(2n-3) + \hdots 5 + 3 + 1}_{\text{sum of first }n-1\text{ odd numbers}} \;=\; n^2 + (n-1)^2$



    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here is a visual "proof" of [1].


    . . $\displaystyle \begin{array}{c} \bullet \end{array} \qquad
    \begin{array}{cc} \bullet & \bullet \\ \circ & \bullet\end{array} \qquad
    \begin{array}{ccc} \bullet & \bullet & \bullet \\ \circ & \circ & \bullet \\ \circ & \circ & \bullet \end{array} \qquad
    \begin{array}{cccc} \bullet & \bullet & \bullet & \bullet \\ \circ & \circ & \circ & \bullet \\ \circ & \circ & \circ & \bullet \\ \circ & \circ & \circ & \bullet \end{array}$

    . . $\displaystyle ^1$ . . . . . $\displaystyle ^{1+3}$ . . . . . $\displaystyle ^{1+3+5}$ . . . . . . .$\displaystyle ^{1+3+5+7}$

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  5. #5
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    1 + 3 + 5 + 7(=n) + 5 + 3 + 1 = 25

    You can also let n = the higher number, and use sum of odds formula : [(n + 1) / 2]^2

    Since the n is added only once:
    SUM = 2[(n + 1) / 2]^2 - n ..... get my drift?
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