After solving all 6 problems, I just remembered a very fast, easy and awesome way to solve other problems like these 6 problems. (Note: You can easily prove my equations!)

Back to our easy equations:

\( 1={ 1 }^{ 2 } \)

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

\( 1+2+3+2+1={ 3 }^{ 2 } \)

...

And:
\( 1={ 1 } \)

\( 1+2=\frac { 2\times (2+1) }{ 2 } \)

\( 1+2+3=\frac { 3\times (3+1) }{ 2 }\)

...

These are very simple right, I think you can prove them easily. But there is a question: Can you calculate quickly this sum: \( 1 + 4 + 7 + ... + 34 = ? \)

My answer is: Yes, you can! But how?

Using this sum above
First, calculate the total terms of this sum:
Numbers of Terms \(= \frac { ( \text{ The last term - The first term} ) } { \text{ The difference between each term}} +1 \).

In this equation, the number of terms is: \( \frac { 34-1 }{ 3 } +1=12 \)

Our first step is completed, now move to the next step To sum this equation, use this: The sum \(= \frac {{ ( \text{The first term + The last term}) \times} {\text {Number of terms }}}{2} \)

So in this equation: The sum is: \(\frac { (34+1)\times 12 }{ 2 } =210\)

This is my solution, and you can use it in all arithmetic series. I hope to discuss more with all of you.

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TopNewestNote: To use latex, you have to place it within the brackets \ ( code \ ), with spaces removed. Also, you don't need to use \quad to space out text. Just leave them outside of the brackets. If you want to include text, I would suggest using \text{ XXX } instead, as this makes it display as per normal, instead of

italicised.I edited the first few paragraphs of your note, to give you an example. You can edit it to see the changes that I made.

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Oh, thank you. I am just a newbie here!

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Looks great! You are a fast learner.

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Thank you for your help!

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