The problem is Find the sum of 4+8+12+16+...+404. I need to not only just add them all and find the answer, but show how to set up the problem, but I am completely lost as to how to even begin. I know the Gauss method for a similar problem, such as 1+2+3+...+100, but I don't know how to make it work for this one. Any advice is greatly appreciated.
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possibly a Gauss method problem
- Thread starter danicat83
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danicat83 said:The problem is Find the sum of 4+8+12+16+...+404. I need to not only just add them all and find the answer, but show how to set up the problem, but I am completely lost as to how to even begin. I know the Gauss method for a similar problem, such as 1+2+3+...+100, but I don't know how to make it work for this one. Any advice is greatly appreciated.
1) Treat the series as Arithmatic Progression with a[sub:u5tr4zcv]0[/sub:u5tr4zcv] = 4 and d = 4
or
2) 4+8+12+16+...+404 = 4(1 + 2 + 3 +.....101)
mathteacher
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Each one is a multiple of 4 so factor out a 4.
4(1 + 2 + . . . 101)
Now use each end value to get a repeating pattern. (I would use 0 + 101, 1 + 100, 2 + 99, . . etc.) You could use 101 + 1 = 102, 2 + 100, but if you do this then you have take into account a middle value which will not have a partner. By including the 0 it also increases the range of integers to 102 integers. Divide the range by 2 to find out how many units of 101 there are.
4( 102/2 (101))
4(1 + 2 + . . . 101)
Now use each end value to get a repeating pattern. (I would use 0 + 101, 1 + 100, 2 + 99, . . etc.) You could use 101 + 1 = 102, 2 + 100, but if you do this then you have take into account a middle value which will not have a partner. By including the 0 it also increases the range of integers to 102 integers. Divide the range by 2 to find out how many units of 101 there are.
4( 102/2 (101))