Hi I really need some guided help with steps on this. The answer was 165 but I don't get how, I am attaching the link because I cannot do the syntax. It is problem B15
http://www.baruch.cuny.edu/sacc/documents/MTH2205_FinalExamv8.pdf
http://www.baruch.cuny.edu/sacc/documents/MTH2205_FinalExamv8.pdf
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the sum of the first n positive integers is n(n+1)/2
in this case 10·11/2 = 55
sum of 3k, for k = 1 to 10
is 3 times the sum of k, for k = 1 to 10
3·55 = 165
in this case 10·11/2 = 55
sum of 3k, for k = 1 to 10
is 3 times the sum of k, for k = 1 to 10
3·55 = 165
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∑3k from k = 1 to 10
= 3∑k from 1 to 10
= 3 [ 1 + 2 + 3 ........10 ]
= 3 (10)(11)/2 ( The sum of first n natural numbers is given by n(n+1)/2 )
= 15*11
= 165
= 3∑k from 1 to 10
= 3 [ 1 + 2 + 3 ........10 ]
= 3 (10)(11)/2 ( The sum of first n natural numbers is given by n(n+1)/2 )
= 15*11
= 165