Key Concepts and Formulas

1. General Term ($T_n$): The ability to find a formula for the $n$-th term of a series is fundamental for summation.
2. Summation Properties: The linearity of summation allows us to split sums: $\sum (a_n \pm b_n) = \sum a_n \pm \sum b_n$. Also, the sum of a constant $c$ over $N$ terms is $\sum_{n=1}^N c = cN$.
3. Sum of a Geometric Progression (GP): The sum of the first $N$ terms of a GP with first term $a$ and common ratio $r$ is $S_N = \frac{a(1-r^N)}{1-r}$ for $r \neq 1$.

Step-by-Step Solution

The problem asks for the sum of the first 20 terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$. Let $S_{20}$ denote this sum.

Step 1: Determine the General Term ($T_n$) of the Series We examine the given terms: $T_1 = 1$, $T_2 = \frac{3}{2}$, $T_3 = \frac{7}{4}$, $T_4 = \frac{15}{8}$, $T_5 = \frac{31}{16}$. Observe the denominators: $1, 2, 4, 8, 16, \dots$. These are powers of 2, specifically $2^0, 2^1, 2^2, 2^3, 2^4, \dots$. For the $n$-th term, the denominator is $2^{n-1}$. Observe the numerators: $1, 3, 7, 15, 31, \dots$. These numbers are one less than powers of 2: $2^1-1, 2^2-1, 2^3-1, 2^4-1, 2^5-1, \dots$. For the $n$-th term, the numerator is $2^n-1$. Thus, the general term $T_n$ is given by: $T_n = \frac{2^n - 1}{2^{n-1}}$ This step is crucial as it allows us to represent any term in the series using a formula dependent on its position $n$.

Step 2: Simplify the General Term ($T_n$) We can rewrite the expression for $T_n$ by dividing each part of the numerator by the denominator: $T_n = \frac{2^n}{2^{n-1}} - \frac{1}{2^{n-1}}$ Using the rule of exponents $\frac{a^m}{a^p} = a^{m-p}$: $T_n = 2^{n - (n-1)} - \frac{1}{2^{n-1}}$ $T_n = 2^1 - \frac{1}{2^{n-1}}$ $T_n = 2 - \frac{1}{2^{n-1}}$ This simplification is important because it decomposes the complex term into a constant part and a part involving a geometric progression, making the summation much easier.

Step 3: Express the Sum of the First 20 Terms ($S_{20}$) We need to find the sum of the first 20 terms, $S_{20} = \sum_{n=1}^{20} T_n$. Substituting the simplified form of $T_n$: $S_{20} = \sum_{n=1}^{20} \left(2 - \frac{1}{2^{n-1}}\right)$ This step formally sets up the summation using the simplified general term.

Step 4: Decompose the Sum using Linearity Using the linearity property of summation, we can split the sum into two parts: $S_{20} = \sum_{n=1}^{20} 2 - \sum_{n=1}^{20} \frac{1}{2^{n-1}}$ This decomposition is a key strategy that transforms a single, potentially difficult sum into two simpler sums that can be evaluated independently.

Step 5: Evaluate the First Sum (Sum of a Constant) The first part is the sum of the constant 2 for 20 terms: $\sum_{n=1}^{20} 2 = 2 \times 20 = 40$ This is a straightforward application of the rule for summing a constant.

Step 6: Evaluate the Second Sum (Sum of a Geometric Progression) The second part is $\sum_{n=1}^{20} \frac{1}{2^{n-1}}$. Let's write out the terms: When $n=1$, term is $\frac{1}{2^{1-1}} = \frac{1}{2^0} = 1$. When $n=2$, term is $\frac{1}{2^{2-1}} = \frac{1}{2^1} = \frac{1}{2}$. When $n=3$, term is $\frac{1}{2^{3-1}} = \frac{1}{2^2} = \frac{1}{4}$. ... When $n=20$, term is $\frac{1}{2^{20-1}} = \frac{1}{2^{19}}$. This is a geometric progression with first term $a=1$, common ratio $r=\frac{1}{2}$, and number of terms $N=20$. Using the GP sum formula $S_N = \frac{a(1-r^N)}{1-r}$: $\sum_{n=1}^{20} \frac{1}{2^{n-1}} = \frac{1 \left(1 - \left(\frac{1}{2}\right)^{20}\right)}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{2^{20}}}{\frac{1}{2}}$ $= 2 \left(1 - \frac{1}{2^{20}}\right) = 2 - \frac{2}{2^{20}} = 2 - \frac{1}{2^{19}}$ This step involves recognizing the series as a GP and applying its sum formula. The simplification of $\frac{2}{2^{20}}$ to $\frac{1}{2^{19}}$ uses exponent rules.

Step 7: Combine the Results to Find the Total Sum Now, we substitute the results from Step 5 and Step 6 back into the equation from Step 4: $S_{20} = (\text{Sum of constant}) - (\text{Sum of GP})$ $S_{20} = 40 - \left(2 - \frac{1}{2^{19}}\right)$ Distributing the negative sign: $S_{20} = 40 - 2 + \frac{1}{2^{19}}$ $S_{20} = 38 + \frac{1}{2^{19}}$ This final step combines the results of the individual summations to obtain the overall sum of the series.

Common Mistakes & Tips

• Exponent Rule Errors: Be careful when simplifying fractions with exponents, such as $\frac{2}{2^{20}}$. Incorrect simplification can lead to a wrong answer.
• Sign Errors: When subtracting an expression in parentheses, ensure the negative sign is distributed correctly to all terms within the parentheses.
• GP Parameters: Double-check the first term ($a$), common ratio ($r$), and the number of terms ($N$) when applying the GP sum formula. An error in any of these will yield an incorrect sum.

Summary

The strategy employed here involves identifying the general term of the series, simplifying it into a form that separates a constant from a geometric progression, and then summing each part using known formulas. The sum of the first 20 terms of the given series is calculated to be $38 + \frac{1}{2^{19}}$.

The final answer is \boxed{38 + {1 \over {{2^{19}}}}}.