- Sep 07, 2015 · sum_(n=1)^50 (2n-1)=1+3+5+7+9+.....=97+99 = 2500 Form a sequence (x_n)= 1, 3, 5, 7, 9, ....., 95, 97, 99. This is the sequence of all the odd numbers between 1 and 99, endpoints included. Clearly this is an arithmetic sequence with common difference d = 2 between terms.
- ok so i know this is wrong.....very wrong. i obviously don't know whats going on here. i have a base array with an output of sum, average, largest test score and lowest test score that i created. i have to add a letter grade to these scores and how many scores are of that letter grade for example if a score above 90 gets entered 3 times it needs to say "the number of students with scores of 90 ...
- The average number each digit could be from 000,000 to 999,999 is (9+0)/2=4.5. Since the average of each number is 4.5 and there are 6 digits the average sum of the digits for a 6 digit number should be 4.5*6=27. There are 1 million numbers from 000,000 to 999,999 so the sum of the digits from 000,000 to 999,999 is 27,000,000.
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