Arithmetic In Arithmetic And Power Of Number

Question

How do i solve arithmetic problem that contain another arithmetic in it and the arithmetic is problem is using power of number?

This is the problem that i run into $(2^1) + (2^1 + 2^2) + (2^1 + 2^2 + 2^3) ... U_n$

Is it possible to find the Un and Sn ?

Hint: consider $U_n + 2$ instead. Calculate a few terms. See what you get. — Arthur, Oct 24 '19 at 17:27
You haven't mentioned $S_n$ anywhere in the post except the final line, how is it defined? Are you wanting $S_n = U_1+U_2+\dots+U_n$? — JMoravitz, Oct 24 '19 at 17:34
i'm sorry i mistaken geometric and arithmetic , the post has been edited — user6668201, Oct 24 '19 at 17:34
@J.W.Tanner i have read the guide many time but i still don't understand how to use it , if possible can you please help me with that , i really appreciate if you help me with that . — user6668201, Oct 24 '19 at 17:39
First and most important step in using mathjax: enclose your equations with dollar signs. $(2^1)+(2^1+2^2)+(2^1+2^2+2^3)$ results in $(2^1)+(2^1+2^2)+(2^1+2^2+2^3)$. That already takes you most of the way for simple equations. — JMoravitz, Oct 24 '19 at 17:40

JMoravitz · Accepted Answer · 2019-10-24T18:20:56.200

0

Hint:

$1=2^1-1$

$1+2 = 2^2-1$

$1+2+4 = 2^3-1$

$1+2+4+8 = 2^4-1$

$1+2+4+8+16 = 2^5-1$

$1+2+4+\dots+2^n = 2^{n+1}-1$

So then, what is $U_n$?

Now, looking at $S_n$, can you see a similar pattern emerging?

Solution:

Given that $1+2+4+\dots+2^n = 2^{n+1}-1$ it follows that $U_n = 2+4+\dots+2^n = 2^{n+1}-2$

Further, we have that $S_n = U_1+U_2+\dots+U_n$

By replacing $U_n$ with what we found earlier and rearranging we have:

$S_n = (2^2-2)+(2^3-2)+\dots+(2^{n+1}-2) = (2^2+2^3+\dots+2^{n+1})-2n$

Recalling that $1+2+4+\dots+2^n+2^{n+1}=2^{n+2}-1$ and by subtracting $1+2$ from each side, we arrive at:

$S_n = (2^{n+2}-4)-2n$

edited Oct 24 '19 at 18:20

answered Oct 24 '19 at 17:38

JMoravitz

After Trying several time , i finally figured that $Un = 2^1 + 2^2 + 2^3 .... 2^n = 2^{n+1} - 2$ , thanks so much !!!! but how do i find the Sn ? – user6668201 Oct 24 '19 at 17:53
@user6668201 Additional mathjax tip, several commands will only pay attention to the first following character. If you want additional characters to be recognized, enclose them in braces. $2^n+1$ produces $2^n+1$ but $2^{n+1}$ produces $2^{n+1}$ – JMoravitz Oct 24 '19 at 17:55
i edited my comment several time before removing the ajax when i stuck with that problem , thanks for helping me so much – user6668201 Oct 24 '19 at 17:57
Now, notice that $S_n = (2^2-2)+(2^3-2)+(2^4-2)+\dots+(2^{n+1}-2)$. Let's rearrange that as $(2^2+2^3+2^4+\dots+2^{n+1}) - 2-2-2-\dots - 2$. The first section should look very similar to $U_n$ and you should be able to spot that pattern in a similar way. The second section is just subtracting $2$ a bunch of times. How many times? – JMoravitz Oct 24 '19 at 17:57
apparently another solution was $S_n = 2^{n+2} - (n+2)(2) $ , but your answer follow more logical way than this solution – user6668201 Oct 24 '19 at 18:33

1 Answers1