Certainly! Let's analyze the series provided: [tex]\( 1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \cdots \)[/tex].
We are asked to sum the first few terms of this series. Let's denote each term of the series as [tex]\( a_n \)[/tex], where [tex]\( n \)[/tex] represents the position of the term in the series. In general, the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex] of this series is given by:
[tex]\[
a_n = \frac{n}{2^{n-1}}
\][/tex]
Now, we will identify the number of terms to sum. We will sum the first 10 terms of the series. Here are the first 10 terms calculated individually:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[
a_1 = \frac{1}{2^{1-1}} = \frac{1}{1} = 1
\][/tex]
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[
a_2 = \frac{2}{2^{2-1}} = \frac{2}{2} = 1
\][/tex]
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[
a_3 = \frac{3}{2^{3-1}} = \frac{3}{4} = 0.75
\][/tex]
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[
a_4 = \frac{4}{2^{4-1}} = \frac{4}{8} = 0.5
\][/tex]
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[
a_5 = \frac{5}{2^{5-1}} = \frac{5}{16} = 0.3125
\][/tex]
6. For [tex]\( n = 6 \)[/tex]:
[tex]\[
a_6 = \frac{6}{2^{6-1}} = \frac{6}{32} = 0.1875
\][/tex]
7. For [tex]\( n = 7 \)[/tex]:
[tex]\[
a_7 = \frac{7}{2^{7-1}} = \frac{7}{64} = 0.109375
\][/tex]
8. For [tex]\( n = 8 \)[/tex]:
[tex]\[
a_8 = \frac{8}{2^{8-1}} = \frac{8}{128} = 0.0625
\][/tex]
9. For [tex]\( n = 9 \)[/tex]:
[tex]\[
a_9 = \frac{9}{2^{9-1}} = \frac{9}{256} = 0.03515625
\][/tex]
10. For [tex]\( n = 10 \)[/tex]:
[tex]\[
a_{10} = \frac{10}{2^{10-1}} = \frac{10}{512} = 0.01953125
\][/tex]
To obtain the sum of these terms, we add them together:
[tex]\[
1 + 1 + 0.75 + 0.5 + 0.3125 + 0.1875 + 0.109375 + 0.0625 + 0.03515625 + 0.01953125
\][/tex]
Adding these values, we get:
[tex]\[
1 + 1 = 2 \\
2 + 0.75 = 2.75 \\
2.75 + 0.5 = 3.25 \\
3.25 + 0.3125 = 3.5625 \\
3.5625 + 0.1875 = 3.75 \\
3.75 + 0.109375 = 3.859375 \\
3.859375 + 0.0625 = 3.921875 \\
3.921875 + 0.03515625 = 3.95703125 \\
3.95703125 + 0.01953125 = 3.9765625
\][/tex]
Therefore, the sum of the first 10 terms of the series [tex]\( 1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \cdots \)[/tex] is:
[tex]\[
\boxed{3.9765625}
\][/tex]
