Find closed formula by changing order of summation: $\sum_{i=1}^ni3^i$

Here is a rather detailed elaboration which might be useful.

We obtain \begin{align*} \color{blue}{\sum_{i=1}^ni3^i}&=\sum_{i=1}^n\left(\sum_{k=1}^i 1\right)3^i\tag{1}\\ &=\sum_{i=1}^n\sum_{k=1}^i 3^i =\sum_{1\leq k\leq i\leq n}3^i =\sum_{k=1}^n\sum_{i=k}^n3^i\tag{2}\\ &=\sum_{k=1}^n\sum_{i=0}^{n-k}3^{i+k}\tag{3}\\ &=\sum_{k=1}^n3^k\cdot\frac{3^{n-k+1}-1}{3-1}\tag{4}\\ &=\frac{1}{2}\sum_{k=1}^n\left(3^{n+1}-3^k\right)\tag{5}\\ &=\frac{n}{2}3^{n+1}-\frac{1}{2}\sum_{k=1}^n3^k\tag{6}\\ &=\frac{n}{2}3^{n+1}-\frac{1}{2}\cdot\left(\frac{3^{n+1}-1}{3-1}-1\right)\tag{7}\\ &=\frac{n}{2}3^{n+1}-\frac{1}{4}3^{n+1}+\frac{3}{4}\tag{8}\\ &\color{blue}{=\frac{n}{4}(2n-1)3^{n+1}+\frac{3}{4}}\tag{9} \end{align*}

Comment:

  • In (1) we represent the factor $i$ as sum.

  • In (2) we multiply out in the left-hand sum and write the index range somewhat more conveniently in the middle sum. We exchange the sums in the right-hand double-sum.

  • In (3) we shift the index of the inner sum to start from $i=0$.

  • In (4) we apply the finite geometric summation formula.

  • In (5) we do some simplifications.

  • In (6) we multiply out and do some simplifications.

  • In (7) we apply the finite geometric summation formula again.

  • In (8) and (9) we do some more simplifications.


Just use a sum of the geometric series.

It's $$\sum_{i=1}^n\frac{3^k(3^{n-k+1}-1)}{3-1}=\frac{n3^{n+1}}{2}-\frac{1}{2}\frac{3(3^n-1)}{2}=\frac{2n3^{n+1}-3^{n+1}+3}{4}.$$


Because the other answer already discusses about computing $\sum_{k=1}^n \sum_{i=k}^n 3^i$, here's another way to compute $\sum_{i=1}^n i3^i$

\begin{eqnarray} S & = & \sum_{i=1}^n i 3^i \\ & = & 1.3^1 + 2.3^2 + 3.3^3 + \cdots + n.3^n \\ 3S & = & \space\space\space\space\space\space\space\space\space\space\space\space 1.3^2 + 2.3^3 + \cdots + (n-1).3^n + n.3^{n+1} \\ \end{eqnarray}

Therefore, on subtracting, we get $$S - 3S = (3^1 + 3^2 + 3^3 + \cdots + 3^n) - n.3^{n+1}$$ $$\implies -2S = 3\left(\frac{3^n-1}{2}\right) - n.3^{n+1} = \frac{(1-2n)3^{n+1}-3}{2}$$ $$\implies S = \frac{(2n-1)3^{n+1}+3}{4}$$

Tags:

Summation

Algebra Precalculus

Closed Form

Exponential Function

Geometric Progressions

Related

If $ \int fg = 0 $ for all compactly supported continuous g, then f = 0 a.e.? $\mathfrak p$ be a prime ideal of a commutative ring $R$ , then $R_\mathfrak p/\mathfrak pR_\mathfrak p$ is the field of fractions of $R/\mathfrak p$? How do I prove these three statements true/false? Union of differences between three sets, where nothing intersects, how is it called? Is this sum $\sum_{k=0}^{n}{n\choose k}\sum_{i=0}^{k}(-1)^i{k\choose i}H_{n+k-i}=H_{2n}$ correct? Little o notation in Taylor's formula : o(1) Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$. Can $\log_a(-b)$ be solved using complex/imaginary numbers? A question about the definition of polynomials. The set of primes of the form $a^2+8b^2$, where $a,b\in\mathbb N_+$ When does a multivariate power series define an entire function? Does there exist a bijective differentiable function $f:\mathbb{R^+}\rightarrow \mathbb{R^+}$, whose derivative is not a continuous function?

Recent Posts