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Mathematics 12 Online
OpenStudy (anonymous):

What’s the sum of the integers from 1 to 300?

OpenStudy (amistre64):

well, we can either count them once, or twice ... which would you perfer

OpenStudy (anonymous):

i dont know how finding integers work :/

OpenStudy (amistre64):

lets take the set:\[1,2,3,4,...,297,298,299,300\] do you agree that there are 300 elements in this set?

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

lets take another set, but define it as: \[300,299,298,297,...,4,3,2,1\] would you agree that there are 300 elements of this set; and that it is just the first one turned around?

OpenStudy (anonymous):

yes :)

OpenStudy (amistre64):

then the sum of adding both sets would be 2 times the amount we need .... so lets add the elements together like this: \[\begin{matrix} &1&2&3&4&...&297&298&299&300\\ +&300&299&298&297&...&4&3&2&1\\ \hline\\ &301&301&301&301&...&301&301&301&301 \end{matrix}\] would you agree that we have 300 elements of 301 each?

OpenStudy (anonymous):

yes i agree

OpenStudy (amistre64):

so, 301 added together 300 times is just multiplication:\[300(301)\]and that is twice the solution. therefore divide it in half:\[\frac{300(301)}{2}\]

OpenStudy (amistre64):

in general: if you have \(n\) elements of an arithmetic sequence, then the sum of them amounts to:\[\frac{n(first+last)}{2}\]

OpenStudy (anonymous):

ok sooo n is 300 ( 1 + 300 ) /2 ?

OpenStudy (anonymous):

or n is 301?

OpenStudy (amistre64):

correct

OpenStudy (anonymous):

oh so n is 300 only? where did the 301 go?

OpenStudy (amistre64):

we have 300 elements; first one is 1, last one is 300 the sum of the set is: 300(1+300)/2

OpenStudy (anonymous):

ph ok so basically 300(301)/2 =45150

OpenStudy (anonymous):

oh*

OpenStudy (amistre64):

correct

OpenStudy (anonymous):

thank you ^_^ ill keep in mind the formula thanks again

OpenStudy (amistre64):

youre welcome

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