Sequence simplification

M

matfanka

New member
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Jul 11, 2020
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6
Hello,
I'm stuck and I can't solve one problem. I've been trying for a few days and have no more ideas.

The sum is defined by: sn=2+8+24+...+n2n
Find simplified expression for the following: tn=n+2(n-1)+4(n-2)+8(n-3)+...+2n-11

I've come to the point:
tn=(n+2n+4n+...+2n)-(2+8+24+...+2nn)=n(-1)-sn=-n-sn
Where I've calculated -1 from the formula for the sum of the infinite geometric sequence. The answer key says: "2n+1-n-2". I don't know how to get there. I'd appreciate someone's help.
 
matfanka said:
Hello,
I'm stuck and I can't solve one problem. I've been trying for a few days and have no more ideas.

The sum is defined by: sn=2+8+24+...+n2n
Find simplified expression for the following: tn=n+2(n-1)+4(n-2)+8(n-3)+...+2n-11

I've come to the point:
tn=(n+2n+4n+...+2n)-(2+8+24+...+2nn)=n(-1)-sn=-n-sn
Where I've calculated -1 from the formula for the sum of the infinite geometric sequence. The answer key says: "2n+1-n-2". I don't know how to get there. I'd appreciate someone's help.
Click to expand...
You state:

I've come to the point: ..........tn=(n+2n+4n+...+2n)-(2+8+24+...+2nn)=n(-1)-sn=-n-sn​

Can you please show us detailed steps of your derivation?
 
tn=n+2n-2+4n-8+8n-24+...+2n-1=(n+2n+4n+...+2n)-(2+8+24+...+2nn)=n*((1)/(1-2))-sn=-n-sn
 
matfanka said:
The sum is defined by: sn=2+8+24+...+n2n
Find simplified expression for the following: tn=n+2(n-1)+4(n-2)+8(n-3)+...+2n-11

I've come to the point:
tn=(n+2n+4n+...+2n)-(2+8+24+...+2nn)=n(-1)-sn=-n-sn
Where I've calculated -1 from the formula for the sum of the infinite geometric sequence. The answer key says: "2n+1-n-2". I don't know how to get there. I'd appreciate someone's help.
Click to expand...

There is no infinite geometric series here. You need the formula for a finite series. (An infinite geometric series with common ratio 2 does not converge!)
 
You're right, thank you! I did it.
Now, the next task is to calculate un=1/2+2/4+3/8+...+n/2n
The key says it's un=tn/2n, I can calculate it then, but I don't understand how I should notice this fact? Where it comes from?
 
matfanka said:
You're right, thank you! I did it.
Now, the next task is to calculate un=1/2+2/4+3/8+...+n/2n
The key says it's un=tn/2n, I can calculate it then, but I don't understand how I should notice this fact? Where it comes from?
Click to expand...
Have you studied:

Sum of geometric series

in Algebra?
 
But where do you think 1/2, 2,4, 3/8, n/2n is a geometric series? It's not formulated by the multiplication by the same number.
I don't understand why un=tn/2n.
 
matfanka said:
Now, the next task is to calculate un=1/2+2/4+3/8+...+n/2n
The key says it's un=tn/2n, I can calculate it then, but I don't understand how I should notice this fact? Where it comes from?
Click to expand...
I don't know. If you haven't been taught it, what have you been taught? Can you list the kinds of series you have learned about? What does tn mean in your class? (I haven't checked your formula yet, but it doesn't look right to me.)

This is not a geometric series, but the formula can be obtained by a similar method to the proof of the formula for geometric series.
 
I've become confused because you didn't state the whole problem at once, but just said "the next task is ..." without saying how things are connected.

As I understand it, you were given a series s_n, with no formula for the sum, then t_n, which you related to s_n. Then, in post #5, you introduced u_n, which apparently you were to relate to the previously stated t_n. (I initially thought this was a new problem entirely, and t_n might be the triangular numbers, for example, though that would be wrong.)

In your work here, you state a formula for s_n without explaining where it came from. Can you explain that?

Then the last three lines, I think you're saying, are from the key, and you don't understand the one statement that u_n = t_n/2^n. I think that probably comes from the observation that t_n = u_n * 2^n. That is, if you multiply each term of u_n by 2^n, you get the same term as in t_n:

u_n * 2^n = (1*2^-1 + 2*2^-2 + 3*2^-3 ...)*2^n = 1*2^{n-1} + 2*2^{n-2} + 3*2^{n-3) ... = t_n.​

This takes some time to see, as they have reversed the order of terms.
 
s_n is given in the task as sn=2+8+24+...+n2n
which is also equal to f(n) in (iii).

Thank you!
 
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