Question about Series

[dxoo]

Problem Statement:
Find the sum of the numbers which are multiples of 8 but not multiples of 6, between 100 and 3100.

My Approach:
I started by writing out the sequence:
[MATH]8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136[/MATH]
I notice the pattern is [MATH]= 8*1, 8*2, 8*4, 8*5, 8*7, 8*8, 8*10, 8*11, 8*13, 8*14, 8*16 ....[/MATH]
But I can't seem to be able to find a formula for the general term.

 

[lev888]

Problem Statement:
Find the sum of the numbers which are multiples of 8 but not multiples of 6, between 100 and 3100.

My Approach:
I started by writing out the sequence:
[MATH]8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136[/MATH]
I notice the pattern is [MATH]= 8*1, 8*2, 8*4, 8*5, 8*7, 8*8, 8*10, 8*11, 8*13, 8*14, 8*16 ....[/MATH]
But I can't seem to be able to find a formula for the general term.

It's easier to go around the mountain.
Can you calculate the sum of multiples of 8?
Now, to get the answer we need to subtract from this number the sum of what?

 

[pka]

Problem Statement:
Find the sum of the numbers which are multiples of 8 but not multiples of 6, between 100 and 3100.

This must be an exercise assigned in a programming class.
S=0
for k=101 to 3099
if \((k \mod 8)=0\text{ and }(k \mod 6)\ne 0) \text{ then } S=S+k\)
next.
print S.

 

[dxoo]

Of course! I'm disappointed in myself for not getting that. Well, here's the rest of my answer that I believe is correct:

Approach:
1. Calculate sum of multiples of 8 between 100 and 3100
2. Calculate sum of multiples of 6 that are also multiples of 8 between 100 and 3100
Subtract 2 from 1.

[MATH] 8(13) = 104\\ 104 + 112 + 120 + ...\\ t_{n} = 104 + (n-1)8\\ 104 + n(-1)8 < 3100\\ n = 375\\ t_{375} = 3096\\ S_{375} = 187.5(104+3096)\\ S_{375} = 600,000\\ [/MATH]
Sequence of numbers divisible by 6 AND 8:
[MATH] 24 + 48 + 72 + 96 + 120 + ...\\ t_{n} = 120 + (n-1)24\\ 120 + (n-1)24 < 3100\\ n = 125\\ t_{125} = 3096\\ S_{125} = 62.5(120+3096)\\ S_{125} = 201,000 [/MATH]
So the sum of numbers that are multiples of 8 between 100 and 3100 is 600,000.
The sum of numbers that are multiples of 6 and 8 between 100 and 3100 is 201,000.

[MATH]600,000 - 201,000 = 399,000[/MATH]
Therefore, the sum of the numbers which are multiples of 8 but not multiples of 6, between 100 and 3100 is 399,000.

Thank you for the idea, lev888.

 

[pka]

It's easier to go around the mountain.
Can you calculate the sum of multiples of 8? Now, to get the answer we need to subtract from this number the sum of what?

The problem with your approach is that there are multiples of eight that are not multiples of six.
\(40\) is a multiple of eight but not a multiple of six.

 

[lev888]

The problem with your approach is that there are multiples of eight that are not multiples of six.
\(40\) is a multiple of eight but not a multiple of six.

Yes, those we want to include in the sum. Not sure what the problem is.