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Old 01-02-2008, 11:41 PM
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Default Sequence & series

Find the sum of the multiply of 3 less than 100. Hence or otherwise find the sum of the number less than 100 which are not multiples of 3.

In this problem what to do? Is that series for 1st portion, 1+3+9+… ?
Where a = 1, d=3n. But how can I find n’s value?

Please help to solve this problem.
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  #2  
Old 01-03-2008, 12:13 AM
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Find the sum of the multiply of 3 less than 100. Hence or otherwise find the sum of the number less than 100 which are not multiples of 3.

In this problem what to do? Is that series for 1st portion, 1+3+9+… ?
Where a = 1, d=3n. But how can I find n’s value?

Please help to solve this problem.
umm, 1 is not a multiple of 3...

we have an arithmetic series here, it is of the form a_n = 3n for n \in \mathbb{N}

we want to add all the terms of this sequence from 3 to 99. now from our formula, we know that 99 is the 33rd term in our sequence. thus we want to find the sum of the first 33 terms of our sequence.

the formula for the sum of the first n terms of an arithmetic sequence is given by:

S_n = \frac {n(a_1 + a_n)}2 where a_1 is the first term, a_n is the nth term, and n is the number of terms we are summing.

so you can find the sum.

we can use this same formula to find the sum of all integers from 1 to 99. then subtract the sum you found before. that will give you the sum of all integers that are not multiples of 3
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Old 01-03-2008, 01:16 AM
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Thank you so much Jhevon. Your solution is too clear to understand. Thank you
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Old 01-03-2008, 03:40 AM
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I’ve another problem,


Prove that the sum of the first 2n multiples of 4 is 4n(2n+1).

What the meaning by first 2n? Is it 2+4+8+ … +2n series?
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Old 01-03-2008, 05:20 AM
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Originally Posted by geton View Post
I’ve another problem,


Prove that the sum of the first 2n multiples of 4 is 4n(2n+1).

What the meaning by first 2n? Is it 2+4+8+ … +2n series?
4 + 8 + 12 + .....
They want the first '2n' terms. You can guess...
1st term = 4,
2nd term = 8,
3nd term = 12,
.
.
.
.
'2n'th term = ??
It is 8n

So,
4+8+12+......+8n = 4(1+2+3....) = 4\frac{2n(2n+1)}{2}
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Old 01-03-2008, 05:49 AM
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I'm stupid

Thank you Isomorphism.
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Old 01-03-2008, 02:20 PM
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I'm stupid
You aren't stupid. You just made a mistake. Don't beat yourself up.

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Old 01-03-2008, 02:23 PM
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Don't beat yourself up.
yes, that's selfish. allow us to do it for you

just kidding!
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