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bumblebee123

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I've been looking at the question for ages and I have no idea where to start. Can anyone help?

question: an arithmetic series has the first term a and common difference 2a. Show that the sum to 2N terms is always equal to four times the sum to N terms

any help would be really appreciated. thanks :)
 
bumblebee123 said:
I've been looking at the question for ages and I have no idea where to start. Can anyone help?

question: an arithmetic series has the first term a and common difference 2a. Show that the sum to 2N terms is always equal to four times the sum to N terms

any help would be really appreciated. thanks :)
Click to expand...
Do you know the expression for the sum of first N-terms of an AP?
 
If the first term is a and the common difference is 2a then the sequence is a, 3a, 5a, 7a, …. In other word it is the sequence of odd numbers, each multiplied by a. In particular, the nth term is (2n-1)a. The sum to N terms is a(1+ 3+ 5+ ….+ (2N+ 1)) and the sum to 2N terms is a(1+ 3+ 5+ …+ (4N-1)).

Adding odd integers, we get 1+ 3= 4, 1+ 3+ 5= 9, 1+ 3+ 5+ 7= 16, 1+ 3+ 5+ 7+ 9=25, ... Do you see a pattern?
 
HallsofIvy said:
If the first term is a and the common difference is 2a then the sequence is a, 3a, 5a, 7a, …. In other word it is the sequence of odd numbers, each multiplied by a. In particular, the nth term is (2n-1)a. The sum to N terms is a(1+ 3+ 5+ ….+ (2N+ 1)) and the sum to 2N terms is a(1+ 3+ 5+ …+ (4N-1)).

Adding odd integers, we get 1+ 3= 4, 1+ 3+ 5= 9, 1+ 3+ 5+ 7= 16, 1+ 3+ 5+ 7+ 9=25, ... Do you see a pattern?
Click to expand...

I'm still a bit confused
 
bumblebee123 said:
I don't think so?
Click to expand...
Yes you do know the the sum of the the first n terms of a sequence which we denote as Sn. I know this because I've seen you use it a few times.

Now a1 = a and d=2a. Treat a and 2a just as you would numbers, just write a and 2a exactly where you would put numbers.

Now they want you to find SN and S2N. Again, instead of numbers they are using letters. Put these letters where you would put numbers.

If you have trouble with this then think a is really 5 BUT write a instead of 5. Think of 2a as 10 but write 2a instead of 10. Think of N as .... and 2N as ...

After doing this if you get stuck please post your work so we can see where you are stuck.
 
Jomo said:
Yes you do know the the sum of the the first n terms of a sequence which we denote as Sn. I know this because I've seen you use it a few times.

Now a1 = a and d=2a. Treat a and 2a just as you would numbers, just write a and 2a exactly where you would put numbers.

Now they want you to find SN and S2N. Again, instead of numbers they are using letters. Put these letters where you would put numbers.

If you have trouble with this then think a is really 5 BUT write a instead of 5. Think of 2a as 10 but write 2a instead of 10. Think of N as .... and 2N as ...

After doing this if you get stuck please post your work so we can see where you are stuck.
Click to expand...

so sN = n/2 [ 2a + ( n -1 )2a ]

sN = n/2 [ 2a + 2an - 2a ]

sN = n/2 [ 2an ]
 
bumblebee123 said:
s2N must = 2 x ( n/2 [ 2an ] )

s2N = n (4an)
Click to expand...
No, S2N\(\displaystyle \neq\)2*SN

If this was true then, for example, S60=60*S1 = 60a1 which is NOT true.

Also don't type n for N.

_____________________________________________________________________________________________________________

a1 = a and d=2a, n=N

Sn = na1 + n(n-1)d/2 Now rewrite the equation to left using N for n, a for a1 and 2a for d.

Let's see what you get.
 
I'll write out SN for you and you write out S2N

Sn = na1 + n(n-1)d/2
SN = Na + N(N-1)2a/2 = Na + N(N-1)a = N2a
 
HallsofIvy said:
If the first term is a and the common difference is 2a then the sequence is a, 3a, 5a, 7a, …. In other word it is the sequence of odd numbers, each multiplied by a. In particular, the nth term is (2n-1)a. The sum to N terms is a(1+ 3+ 5+ ….+ (2N+ 1)) and the sum to 2N terms is a(1+ 3+ 5+ …+ (4N-1)).

Adding odd integers, we get 1+ 3= 4, 1+ 3+ 5= 9, 1+ 3+ 5+ 7= 16, 1+ 3+ 5+ 7+ 9=25, ... Do you see a pattern?
Click to expand...
The "pattern" I was hoping you would see is that each sum is a square! The sum of the first two odd numbers, 1+ 3, is 2 squared, the sum of the first three odd numbers is 3 squared, the sum of the first four odd numbers is 4 squared, and the sum of the first five odd numbers is 5 squared. It should not be too difficult to conjecture that "the sum of the first "n" odd numbers is n squared". It would be a good exercise to prove that "by induction".
 
Hallsof Ivy, I saw your post and even knew what you wrote. The OP seems to need help with the basics and thought that what I was doing with the OP would be helpful. After all the OP said that S2N = 2SN. I assure you that at the end I was going to ask the OP to look at your post and let him see the easiest why to do this problem. Also a proof by induction for a student in arithmetic might be too ambitious.
 
bumblebee123 said:
question: an arithmetic series has the first term a and common difference 2a. Show that the sum to 2N terms is always equal to four times the sum to N terms
Click to expand...
\(\displaystyle \begin{align*}\sum\limits_{k = 1}^{2N} {{a_k}} &= \sum\limits_{k = 1}^{2N} {[a + (k - 1)(2a)]} \\
&= \sum\limits_{k = 1}^{2N} a + 2a\sum\limits_{k = 1}^{2N} {(k - 1)} \\& = 2aN + 2a\left[ {\frac{{(2N - 1)(2N)}}{2}} \right]\\& = 2aN + 4{N^2}a - 2aN\\& = 4a{N^2}\end{align*}\)________ NOW________ \(\displaystyle \begin{align*}\sum\limits_{k = 1}^N {{a_k}} &= \sum\limits_{k = 1}^N {[a + (k - 1)(2a)]} \\& = \sum\limits_{k = 1}^N a + 2a\sum\limits_{k = 1}^N {(k - 1)} \\& = aN + 2a\left[ {\frac{{(N - 1)(N)}}{2}} \right]\\& = aN + {N^2}a - aN\\& = a{N^2}\end{align*}\)

What do you have to say now?
 
Jomo said:
I'll write out SN for you and you write out S2N

Sn = na1 + n(n-1)d/2
SN = Na + N(N-1)2a/2 = Na + N(N-1)a = N2a
Click to expand...

I'm not sure what formula is being used, I don't think I've ever used that one. I usually use: Sn = n/2 [ 2a + ( n - 1 )d ]

using that, is S2N = 2N/2 [ 2a + ( 2N -1 )2a ]

sorry, this is really difficult for me to understand
 
pka said:
\(\displaystyle \begin{align*}\sum\limits_{k = 1}^{2N} {{a_k}} &= \sum\limits_{k = 1}^{2N} {[a + (k - 1)(2a)]} \\
&= \sum\limits_{k = 1}^{2N} a + 2a\sum\limits_{k = 1}^{2N} {(k - 1)} \\& = 2aN + 2a\left[ {\frac{{(2N - 1)(2N)}}{2}} \right]\\& = 2aN + 4{N^2}a - 2aN\\& = 4a{N^2}\end{align*}\)________ NOW________ \(\displaystyle \begin{align*}\sum\limits_{k = 1}^N {{a_k}} &= \sum\limits_{k = 1}^N {[a + (k - 1)(2a)]} \\& = \sum\limits_{k = 1}^N a + 2a\sum\limits_{k = 1}^N {(k - 1)} \\& = aN + 2a\left[ {\frac{{(N - 1)(N)}}{2}} \right]\\& = aN + {N^2}a - aN\\& = a{N^2}\end{align*}\)

What do you have to say now?
Click to expand...

I'm a bit confused because I don't think I've ever used that formula
 
bumblebee123 said:
I'm not sure what formula is being used, I don't think I've ever used that one. I usually use: Sn = n/2 [ 2a + ( n - 1 )d ]

using that, is S2N = 2N/2 [ 2a + ( 2N -1 )2a ]

sorry, this is really difficult for me to understand
Click to expand...
Continue by simplifying what you have.
 
bumblebee123 said:
I'm a bit confused because I don't think I've ever used that formula
Click to expand...
What formula?
Surely you know that \[\sum\limits_{k = 1}^N k = \frac{{N(N + 1)}}{2}\]
A small variation gives same result: \[\sum\limits_{k = 0}^N k = \frac{{N(N + 1)}}{2}\]
And also \[\sum\limits_{k = 1}^{N-1} k = \frac{{(N-1)(N)}}{2}\]
 
bumblebee123 said:
I'm not sure what formula is being used, I don't think I've ever used that one.
I usually use: Sn = n/2 [ 2a + ( n - 1 )d ]
using that, is S2N = 2N/2 [ 2a + ( 2N -1 )2a ]
sorry, this is really difficult for me to understand
Click to expand...
Pka is starting with the exotic "representative formula" with the scary
"sum symbol" (the one that scares most students!).
The formula you use is correct (same as Pka's):
Sum of 1st n terms = n/2 [2a + ( n - 1 )d ]
Usually shown this way: n[2a + d(n - 1)] / 2

So we have for n terms:
n[2a + d(n - 1)] / 2
= n(2a + dn - d) / 2 ; substitute d = 2a
= n(2a + 2an - 2a) / 2
= (2an + 2an^2 - 2an) / 2
= 2an^2 / 2
= an^2

Now for 2n terms; we start similarly with:
n[2a + d(n - 1)] / 2
= n(2a + dn - d) / 2 ; substitute d = 2a and n = 2n:
= 2n(2a + 4an - 2a) / 2
= n(2a + 4an - 2a)
= 2an + 4an^2 - 2an
= 4an^2

It always helps if you make up a simple case; as example:
3, 9, 15 ,21, 27, 33
a = 3
d = 2a = 6
n = 3 (so 2n = 6)

3 + 9 + 15 = 27
3 + 9 + 15 + 21 + 27 + 33 = 108
108 / 27 = 4
 
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