Question

HeyyThe sum of the first ten terms abeg1, 3/2, 7/4, 15/8, ...

Answer 1

First, recall the following formulas:

$\begin{gathered} \sum ^k_{n\mathop=1}c=ck \\ \\ \sum ^k_{n\mathop=1}r^{n-1}=\frac{1-r^k}{1-r} \end{gathered}$

Next, find an expression for the n-th term of the given sequence. Notice that the given terms can be rewritten as follows:

$\begin{gathered} 1=2-1=2-\frac{1}{2^0} \\ \frac{3}{2}=2-\frac{1}{2}=2-\frac{1}{2^1} \\ \frac{7}{4}=2-\frac{1}{4}=2-\frac{1}{2^2} \\ \frac{15}{8}=2-\frac{1}{8}=2-\frac{1}{2^3} \end{gathered}$

Then, the n-th term is given by the expression:

$a_n=2-\frac{1}{2^{n-1}}$

This expression accurately predicts the first four terms of the given sequence:

$\begin{gathered} a_1=2-\frac{1}{2^{1-1}}=2-\frac{1}{2^0}=2-\frac{1}{1}=2-1=1 \\ a_2=2-\frac{1}{2^{2-1}}=2-\frac{1}{2^1}=2-\frac{1}{2}=\frac{4}{2}-\frac{1}{2}=\frac{3}{2} \\ \ldots \end{gathered}$

Then, the sum of the first ten terms is:

$\begin{gathered} \sum ^{10}_{n\mathop=1}a_n=\sum ^{10}_{n\mathop{=}1}(2-\frac{1}{2^{n-1}}) \\ =\sum ^{10}_{n\mathop{=}1}2-\sum ^{10}_{n\mathop{=}1}\frac{1}{2^{n-1}} \\ =10\cdot2-\sum ^{10}_{n\mathop{=}1}(\frac{1}{2^{}})^{n-1} \\ =20-\frac{1-(\frac{1}{2})^{10}}{1-(\frac{1}{2})} \\ =20-\frac{1-\frac{1}{2^{10}}}{1-\frac{1}{2}} \\ =20-\frac{(\frac{2^{10}-1}{2^{10}})}{(\frac{1}{2})} \\ =20-2(\frac{2^{10}-1}{2^{10}}) \\ =20-\frac{2^{10}-1}{2^9} \\ =20-\frac{1024-1}{512} \\ =20-\frac{1023}{512} \\ =\frac{20\cdot512-1023}{512} \\ =\frac{10240-1023}{512} \\ =\frac{9217}{512} \\ =18.001953125 \end{gathered}$

Therefore, the sum of the first ten terms of the sequence is equal to 9217/512, which in decimal notation is equal to 18.001953125.