Efficient Evaluation of Sigma Notation: Solving for the Sum of a Series

[mateomy]

The problem posed is:
Evaluate
<br /> \sum_{i=1}^{n} (2i + 2^i)<br />

I know that I can break the summation down to this:

<br /> 2\sum_{i=1}^{n} i\, +\, 2\sum_{i=1}^{n}1^i<br />

and then after using some Fundamental Theorems...

<br /> =2\Bigg(\frac{n(n+1)}{2}\Bigg) + 2^n<br />

I can't seem to get it to look like my answer key which is...

<br /> 2^{n+1} + n^2 + n-2<br />

Clearly I am doing something wrong, I know I can expand my last step but when I do, it doesn't look anything close to what the book is showing me. Particularly, where are they getting the n-2 ? Where is my step wrong? THanks in advance for any help.

 

[Mark44]

The problem posed is:
Evaluate
<br /> \sum_{i=1}^{n} (2i + 2^i)<br />

I know that I can break the summation down to this:

<br /> 2\sum_{i=1}^{n} i\, +\, 2\sum_{i=1}^{n}1^i<br />

The first summation is correct, but not the second. 2ⁿ \neq 2 * 1ⁿ.

and then after using some Fundamental Theorems...

<br /> =2(\frac{n(n+1)}{2}) + 2^n<br />

I can't seem to get it to look like my answer key which is...

<br /> 2^{n+1} + n^2 + n-2<br />

Clearly I am doing something wrong, I know I can expand my last step but when I do, it doesn't look anything close to what the book is showing me. Particularly, where are they getting the n-2 ? Where is my step wrong? THanks in advance for any help.

 

[mateomy]

Am I just supposed to expand it from 1 to n then? (the second summation) I can't find anything in this chapter that deals with integers to a variable power...(thanks btw)

 

[Mark44]

\sum 2^i is a finite geometric series, right? You should have seen problems that deal with series like this.

 

[mateomy]

Okay, so I figured out the geometric sequence part of the problem basically being 2^i which is the form of

<br /> \frac{1-2^{n+1}}{1-2}<br />

Is that right? I don't think that it is because I am missing a factor of 2 in there, I believe? (or not?)

 

[Mark44]

You're off by a little bit. For the geometric series you should get
\frac{2-2^{n+1}}{1-2}

which is the same as 2ⁿ⁺¹ - 2.

Notice that the series starts with i = 1, not i = 0, so the first term of the geometric series is 2, not 1.

 

[mateomy]

Thank you so much, this helped a lot.

 

[mateomy]

One more question...

Im confused as to why the numerator is 2-2^{n+1} specifically the individual 2 in the front. My initial geometric sequence (see above) is -numerator only- 1-2^{n+1}, is that one just a default and you're supposed to be putting the outcome of the first term in front?
Again, sorry for the 20 questions, but the section in my book says absolutely nothing on geometric sequence with regards to summation, I had to dig that equation out of Google. THanks again, this is really really helping.

 

[Mark44]

The geometric series part is
\sum_{i=1}^{n}2^i

If we expand this, we get
S_n = 2 + 4 + ... + 2ⁿ
2S_n = 4 + 8 + ... + 2ⁿ + 2^{n + 1}

so 2S_n - S_n = 2^{n + 1} - 2
==> Sn = 2^{n + 1} - 2

or,
\sum_{i=1}^{n}2^i = 2^{n + 1} - 2

 

[mateomy]

Perfect, thank you very much.