What’s the sum of the integers from 1 to 300?
well, we can either count them once, or twice ... which would you perfer
i dont know how finding integers work :/
lets take the set:\[1,2,3,4,...,297,298,299,300\] do you agree that there are 300 elements in this set?
yes
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?
yes :)
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?
yes i agree
so, 301 added together 300 times is just multiplication:\[300(301)\]and that is twice the solution. therefore divide it in half:\[\frac{300(301)}{2}\]
in general: if you have \(n\) elements of an arithmetic sequence, then the sum of them amounts to:\[\frac{n(first+last)}{2}\]
ok sooo n is 300 ( 1 + 300 ) /2 ?
or n is 301?
correct
oh so n is 300 only? where did the 301 go?
we have 300 elements; first one is 1, last one is 300 the sum of the set is: 300(1+300)/2
ph ok so basically 300(301)/2 =45150
oh*
correct
thank you ^_^ ill keep in mind the formula thanks again
youre welcome
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