#Help with a olympiad question.

46 messages · Page 1 of 1 (latest)

jagged fog
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Show that the number 2(1+2+3+4+5+...+2021+2022)+2023 is a perfect square.

tulip bone
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arithmetic progression

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sum of AP has a formula

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$$\frac{n}{2}(a+l)$$

mighty plankBOT
tulip bone
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n -> no. of terms
a -> first term
l -> last term

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we'll calculate this first with the above formula

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,w 2022/2 * (1+2022)

tulip bone
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,calc sqrt(2*2045253 + 2023)

mighty plankBOT
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Result:

2023
tulip bone
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hence perfect square

jagged fog
# mighty plank **AKG**

Oh. So this formula works for all of the exercises that are needed to find the sum of a large number of numbers in the parenthesis?

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Like (1+3+5+7+9+...+366)?

tulip bone
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there's another

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$$\frac{n}{2}(2a+(n-1)d)$$

mighty plankBOT
tulip bone
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n -> of terms
a -> first term
d -> common difference between 2 terms

jagged fog
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Is there another formula for a common difference between 3 terms or 4?

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Or do I just substitute d with the common difference in each case

tulip bone
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$$T_n = a + (n-1)d$$

mighty plankBOT
tulip bone
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Tn -> nth term
a -> first term
d -> common difference

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these 3 are enough

jagged fog
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Oh okay

jagged fog
tulip bone
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if we talk about T3 it means 3rd term which is 10 here

jagged fog
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Oh, okay

jagged fog
tulip bone
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let's have a look at perfect square's definition

jagged fog
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Oh, yeah, that's what I wanted to say, my bad.

tulip bone
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,w perfect square

tulip bone
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for example: 25 is a perfect square because it's square root is a whole number ie 5

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but 2 is not a perfect square it's square root is 1.4142135624... which is not a whole number

jagged fog
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I get it, but this number is too large for me to know. 25 and other numbers like 36, 49, 64,81, 100,121 are learned

tulip bone
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example:

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324's square root will be multiplication of the common pairs which is 2 and 3s above
2*3*3 => 18

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now there's a chance that we won't get common pair for example 8:

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we got a pair but (purple line) but still one number is left alone (green circle), so 8 is not a perfect square