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i need help to solve a question that say :

Sn the sum of the first n odd numbers

1 - Formalize this sum: (Give mathematical expression)?

2 - Calculate this sum?

how to solve that ? :/

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    $\begingroup$ Compute the first few case. $1=1$, $1+3 = 4$, $1+3+5=9$. Keep going until you can guess the pattern. $\endgroup$
    – GEdgar
    Jul 31, 2018 at 21:36
  • $\begingroup$ Do you know the formula for $\sum_{i=1}^{n} i$? $\endgroup$
    – user
    Jul 31, 2018 at 21:38
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    $\begingroup$ @Adel It could be useful see your full work for the solution here in the OP. Bye $\endgroup$
    – user
    Jul 31, 2018 at 21:51
  • $\begingroup$ @gimusi yep i know this formula, but how to calculate the odd numbers $\endgroup$
    – Adel
    Jul 31, 2018 at 22:07

3 Answers 3

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HINT

  • consider the sum of the first $2n$ numbers, that is

$$\sum_{k=1}^{2n} k=1+2+3+\ldots+(2n-1)+2n$$

  • subtract the sum of the first $n$ even numbers picking out a factor 2

$$2+4+6+\ldots+(2n-2)+2n=2\sum_{k=1}^{n} k$$

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  • $\begingroup$ i cant understand, can u explain ! $\endgroup$
    – Adel
    Jul 31, 2018 at 22:06
  • $\begingroup$ The idea is to sum all the even and odd numbers since we know the formula, that is for example to obtain the sum of the first 3 odd numbers, that is up to 5 we calculate: $$1+3+5=(1+2+3+4+5+6)-(2+4+6)=(1+2+3+4+5+6)-2(1+2+3)$$ Can you see the method now? $\endgroup$
    – user
    Jul 31, 2018 at 22:09
  • $\begingroup$ yes, now its very clear thank u :) $\endgroup$
    – Adel
    Jul 31, 2018 at 22:10
  • $\begingroup$ @Adel Well done. You are welcome! If you want to show your final expression here above in the OP, I can get a look to it. Bye $\endgroup$
    – user
    Jul 31, 2018 at 22:11
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There is a very neat pictorial proof of the solution dated as far back as the Pythagorean schools who considered every number to have a shape apparently.

Try drawing the odd numbers as "L" shapes of dots,

i.e. \begin{array}{cc} \blacksquare \\ \\ \end{array} \begin{array}{cc} \blacksquare \\ \blacksquare & \blacksquare\\ \end{array} \begin{array}{cc} \blacksquare \\ \blacksquare \\ \blacksquare & \blacksquare & \blacksquare\\ \end{array}for 1, 3, and 5 and so on, then see what happens when you put them together.

p.s. apologies for my LaTeX, if someone can make it look prettier by all means go ahead and edit. Thanks

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  • $\begingroup$ +1: good answer. Suggestion about the $\LaTeX$: \color seems to work with MathJax {\color{blue}\blacksquare} etc. gives me: $\color{blue}\blacksquare$, $\color{red}\blacksquare$, $\color{green}\blacksquare$, $\ldots$ $\endgroup$
    – Rob Arthan
    Jul 31, 2018 at 22:37
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Here's the $A=B$ solution: $S_1 = 1 = 1 = 1^2$, $S_2 = 1 + 3 = 4 = 2^2$, $S_3 = 1 + 3 + 5 = 9 = 3^2$. Taking differences, we see that the formula must be a quadratic in $n$; three values of a quadratic determine the quadratic: we must have $S_n = n^2$ for all $n$.

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