# Thread:

1. ## Number pattern

Hi, The question I have is:

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

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

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

but I'm not sure how to show all of this generally, so that it works for every case . Any ideas?

2. 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 ?

3. 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

4. Hello, cozza!

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

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

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

and show that your generalisation always holds true.

Look at the left side:

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

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

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

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

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

Therefore, we have:

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

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

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

. . $\begin{array}{c} \bullet \end{array} \qquad<br /> \begin{array}{cc} \bullet & \bullet \\ \circ & \bullet\end{array} \qquad<br /> \begin{array}{ccc} \bullet & \bullet & \bullet \\ \circ & \circ & \bullet \\ \circ & \circ & \bullet \end{array} \qquad<br /> \begin{array}{cccc} \bullet & \bullet & \bullet & \bullet \\ \circ & \circ & \circ & \bullet \\ \circ & \circ & \circ & \bullet \\ \circ & \circ & \circ & \bullet \end{array}$

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

5. 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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