Thread: Number pattern

Hi, The question I have is:

Express a single generality that generalises all three related observations that



and show that your generalisation always holds true.

I know that the answer (i.e. RHS of =) is always (supposed to say no. of different terms)

I also know that LHS is , where a=1


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

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 ?

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

Hello, cozza!

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

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

. .

and show that your generalisation always holds true.

Look at the left side:

. .



In general, we have: .

This is a arithmetic series with:
. . first term , common difference , and terms.

Its sum is: .

Hence, the sum of the first odd numbers is .[1]



Therefore, we have:

. .



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


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


. .

. . . . . . . . . . . . . . . . . . .

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?