0

Find the sum to $n$ terms of the series $$3 × 2+5 × 2^2+7 × 2^3+\cdots$$

My Attempt:

The given series: $$3 × 2+5 × 2^2+7 × 2^3+\cdots$$

$n$th term of $3,5,7,\cdots$ is $2n+1$,

$n$th term of $2, 2^2, 2^3,\cdots$ is $2^n$.

So, the $n$th term of the series is: $$t_n=(2n+1) × 2^n.$$

Now, the sum to $n$ terms of the series is: $$S=\sum t_n=\sum (2n+1) × 2^n=\sum 2n × 2^n+\sum 2^n.$$

jimjim
  • 9,675
pi-π
  • 7,416
  • 1
    You still have to compute those two sums. What is your question? – saulspatz Apr 09 '18 at 05:47
  • 1
    @saulspatz, My question is how to compute those sums? – pi-π Apr 09 '18 at 05:49
  • 1
    There are formulas you can and should have learned. The one on the right is a geometric series of the form $\sum\limits_{k=1}^n a^k$. The one on the left is twice the result of a related summation of the form $\sum\limits_{k=1}^n k\cdot a^k$. For the first, see this wiki page. For the second, see this related question. – JMoravitz Apr 09 '18 at 05:54
  • 1
    As an aside, be careful not to use variables for multiple different purposes within the same post. You were asked to find the sum of the first $n$ terms, $n$ here having the role of the total number of terms in the series. Later in the same post you used $n$ as an indexing variable, $n$ referring to the position of an arbitrary term in the sequence. This should not have happened. You should either change the $n$ in the first line to be "find the sum to $N$ terms of the series" or you should change the $n$ in the later parts to some other common index variable such as $k$. – JMoravitz Apr 09 '18 at 05:59
  • 3
    To the downvoters: The problem statement is clear, and they have shown their own attempt up to, I assume, the point where they are stuck. What more do you expect from a question post? Do you really need the person to explicitly ask "Where do I go from here?" to understand what he wants? – Arthur Apr 09 '18 at 06:05
  • @Arthur: ignoring the "attempt", I would look for the source of the problem and some context to explain why it is of mathematical interest. There are infinitely many series we could look at, so not all of them can be interesting - why is this one worth looking at? Perhaps because it was found somewhere - where? – Carl Mummert Apr 09 '18 at 11:19
  • 1
    @CarlMummert That's unreasonably high, looking at the majority of question posts on this site. While that might add to the question quality, I really don't think that its lacking warrants a downvote. Also, requiring people to write it wouldn't give anything to the quality of questions like these, since a lot of times it's just "I had this exercise in my textbook, and I got stuck", or possibly "I just thought of this cool problem on my own, and I couldn't solve it". The ones who post here because they genuinely encountered a problem in some relevant context are in the minority here, i think. – Arthur Apr 09 '18 at 11:35

3 Answers3

1

Since you are supposed to compute the sum of the firs $n$ terms, it would be better to use a different index of summation. We need to compute$$ 2\sum_{k=1}^n{k2^{k}}+\sum_{k=1}^n{2^k}$$ The trick for the first is to notice that it looks something like the derivative of $\sum{x^k}$ evaluated at $x=2$. Let's rewrite it: $$2\sum_{k=1}^n{k2^{k}}+\sum_{k=1}^n{2^k}=4\sum_{k=1}^n{k2^{k-1}}+\sum_{k=1}^n{2^k} $$

The second sum is a geometric series, which I presume you know how to deal with. For the first, write $$f(x)=4\sum_{k=1}^n{x^k}$$ Then you can see that the first sum is just $f'(2).$

You take over now.

saulspatz
  • 53,131
1

We have $$\sum_{k=1}^n2^k=2^{n+1}-2$$ This should be known to you as I doubt you were given this exercise without having gone through geometric series first.

Next, we tackle the other sum, which I will call $S$, for convenience. We have $$ S=\sum_{k=1}^n 2k2^k\\ =\sum_{k=1}^n \left(2\cdot2^k+2(k-1)2^k\right)\\ =\sum_{k=1}^n2\cdot2^k+\sum_{k=1}^n2(k-1)2^k $$ The first term here is twice the series from the first paragraph, so it is equal to $2(2^{n+1}-2)=2^{n+2}-4$. For the second term, index renaming gives $$ \sum_{k=1}^n2(k-1)2^k\\ =\sum_{k=0}^{n-1}2k2^{k+1}\\ =\sum_{k=0}^{n-1}2k2^k\cdot 2\\ =2\cdot0\cdot 2^0\cdot2+2\left(\sum_{k=1}^{n}2k2^k\right)-2\cdot n\cdot 2^n\cdot 2 $$ The first term is $0$, so we can remove that. The expression inside the brackets is just $S$ again. The two factors $2$ in the last term can be absorbed into the exponent, so that it equals $n2^{n+2}$. So, we can insert this into our original calculation: $$ S=\sum_{k=1}^n2\cdot2^k+\sum_{k=1}^n2(k-1)2^k\\ =2^{n+2}-4+2S-n2^{n+2}\\ S=(n-1)2^{n+2}+4 $$ So the final answer is $$ \sum_{k=1}^n(2k+1)2^k=2^{n+1}-2+(n-1)2^{n+2}+4\\ =(2n-1)2^{n+1}+2 $$

Arthur
  • 199,419
0

$$ \begin{array}{l} Q:the\ sum\ of\sum_{i=1}^{n} (2i+1)*2^i \end{array} $$ $$ \begin{array}{l} \sum_{i=1}^{n} (2i+1)*2^i\\ =\sum_{i=1}^{n} i*2^{i+1}+2^i\\ =\sum_{i=1}^{n} i*2^{i+1}+2^{n+1}-2\\ \\ let \\ \\ S=\sum_{i=1}^{n} i*2^{i+1}\\ then\\ \\ \frac{S}{2}=\sum_{i=1}^{n} i*2^{i}\\ \frac{S}{2}-\sum_{i=1}^{n}2^i+n*2^{n+1}=S \\ S=n*2^{n+2}-\sum_{i=2}^{n+1}2^i\\ =n*2^{n+2}-2^{n+2}+4\\ \\ ANS=2+n*2^{n+2}-2^{n+1} \end{array} $$

abuka_benka
  • 1