Math Problem Solver

The sum of the sequence is $\displaystyle 1022$

This is a geometric sequence.

The nth term in this sequence is given by the formula:

Explicit Formula: $$a_n=2^{n}$$

Recursive Formula: $$a_n=2 a_{n-1}, \text{where } a_1=2^{1}$$

Summation Formula:

Option 1:

$$S_n=\frac{a_1 (r^n - 1)}{r - 1}$$ where $r\neq1$, $r$ is the common ratio, $a_1$ is the 1^st term, and $n$ is the is the term number.

Option 2:

$$\begin{aligned} S_n&=\sum_{i=1}^{n} a_{i} \\ &=\sum_{i=1}^{n} 2^{i} \\ &= \sum_{i=0}^{n - 1} 2^{i + 1} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k + p}^{m + p} f{\left(n - p \right)} = \sum_{n=k}^{m} f{\left(n \right)}}\\&= \sum_{i=0}^{n - 1} 2 \cdot 2^{i}\\&= 2 \sum_{i=0}^{n - 1} 2^{i} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= \frac{2 \cdot \left(1 - 2^{1 + n - 1}\right)}{1 - 2} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} a^n = \frac{1-a^{m+1}}{1-a}}\\&=2 \cdot 2^{n} - 2 \end{aligned}$$where $n$ is the is the term number.

Use the explicit formula to find the index of the final term in the sequence.$$\begin{aligned}2^{9} &= 2^{n} \\ n &= 9\end{aligned}$$Then use the summation formula to solve.$$\begin{aligned}S_n &= 2 \cdot 2^{n} - 2 \\S_{9} &= -2 + 2 \cdot 2^{9}\\S_{9} &= 1022\end{aligned}$$