# Find the sum to $n$ terms of the series

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: $$S=\sum t_n=\sum (2n+1) × 2^n=\sum 2n × 2^n+\sum 2^n.$$

• You still have to compute those two sums. What is your question? – saulspatz Apr 9 '18 at 5:47
• @saulspatz, My question is how to compute those sums? – pi-π Apr 9 '18 at 5:49
• 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 9 '18 at 5:54
• 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 9 '18 at 5:59
• 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 9 '18 at 6:05

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 $$\sum_{k=1}^n(2k+1)2^k=2^{n+1}-2+(n-1)2^{n+2}+4\\ =(2n-1)2^{n+1}+2$$
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 $f'(2).$