Sure, let's work through the sum of the given series:
[tex]\[
1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \cdots
\][/tex]
We are summing this series up to 10 terms.
The general term of the series can be written as:
[tex]\[
\frac{n}{2^{n-1}}
\][/tex]
where [tex]\( n \)[/tex] represents the position in the series.
We can write the sum of the first 10 terms as:
[tex]\[
\sum_{n=1}^{10} \frac{n}{2^{n-1}}
\][/tex]
Let's compute this sum term by term:
1. For [tex]\(n = 1\)[/tex]:
[tex]\[
\frac{1}{2^{1-1}} = \frac{1}{2^0} = 1
\][/tex]
2. For [tex]\(n = 2\)[/tex]:
[tex]\[
\frac{2}{2^{2-1}} = \frac{2}{2^1} = 1
\][/tex]
3. For [tex]\(n = 3\)[/tex]:
[tex]\[
\frac{3}{2^{3-1}} = \frac{3}{2^2} = \frac{3}{4} = 0.75
\][/tex]
4. For [tex]\(n = 4\)[/tex]:
[tex]\[
\frac{4}{2^{4-1}} = \frac{4}{2^3} = \frac{4}{8} = 0.5
\][/tex]
5. For [tex]\(n = 5\)[/tex]:
[tex]\[
\frac{5}{2^{5-1}} = \frac{5}{2^4} = \frac{5}{16} = 0.3125
\][/tex]
6. For [tex]\(n = 6\)[/tex]:
[tex]\[
\frac{6}{2^{6-1}} = \frac{6}{2^5} = \frac{6}{32} = 0.1875
\][/tex]
7. For [tex]\(n = 7\)[/tex]:
[tex]\[
\frac{7}{2^{7-1}} = \frac{7}{2^6} = \frac{7}{64} = 0.109375
\][/tex]
8. For [tex]\(n = 8\)[/tex]:
[tex]\[
\frac{8}{2^{8-1}} = \frac{8}{2^7} = \frac{8}{128} = 0.0625
\][/tex]
9. For [tex]\(n = 9\)[/tex]:
[tex]\[
\frac{9}{2^{9-1}} = \frac{9}{2^8} = \frac{9}{256} = 0.03515625
\][/tex]
10. For [tex]\(n = 10\)[/tex]:
[tex]\[
\frac{10}{2^{10-1}} = \frac{10}{2^9} = \frac{10}{512} = 0.01953125
\][/tex]
Adding up all these terms gives us:
[tex]\[
1 + 1 + 0.75 + 0.5 + 0.3125 + 0.1875 + 0.109375 + 0.0625 + 0.03515625 + 0.01953125
\][/tex]
Summing these values provides the final result:
[tex]\[
3.9765625
\][/tex]
So 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]\( 3.9765625 \)[/tex].
