Sum $\sum(2n+1) × 2^n$

Question

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

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

Answer 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 $f'(2).$

You take over now.

Answer 3

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