Sequence & series

geton · #1 $Old$ 01-02-2008, 11:41 PM

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.

Jhevon · #2 $Old$ 01-03-2008, 12:13 AM

Quote:

Originally Posted by geton $View Post$

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

geton · #3 $Old$ 01-03-2008, 01:16 AM

Thank you so much Jhevon. Your solution is too clear to understand. Thank you

geton · #4 $Old$ 01-03-2008, 03:40 AM

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?

Isomorphism · #5 $Old$ 01-03-2008, 05:20 AM

Quote:

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....) $= 4\frac{2n(2n+1)}{2}$