Key Concepts and Formulas To solve this problem, we will utilize the following fundamental identities involving binomial coefficients, derived from the Binomial Theorem:

1. Sum of all Binomial Coefficients: The sum of all binomial coefficients for a given $n$ is given by: $\sum_{r=0}^{n} {}^{n}C_r = {}^{n}C_0 + {}^{n}C_1 + \dots + {}^{n}C_n = 2^n$

   • Why this works: This identity is a direct consequence of the binomial expansion of $(1+x)^n = \sum_{r=0}^{n} {}^{n}C_r x^r$. If we substitute $x=1$ into this expansion, we get $(1+1)^n = \sum_{r=0}^{n} {}^{n}C_r (1)^r$, which simplifies to $2^n = \sum_{r=0}^{n} {}^{n}C_r$.

2. Sum involving $r \cdot {}^{n}C_r$: The sum of $r$ times the binomial coefficient ${}^{n}C_r$ is given by: $\sum_{r=0}^{n} r \cdot {}^{n}C_r = n \cdot 2^{n-1}$

   • Why this works: This identity can be elegantly derived by differentiation. Consider the binomial expansion: $(1+x)^n = \sum_{r=0}^{n} {}^{n}C_r x^r$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(1+x)^n = \frac{d}{dx}\left(\sum_{r=0}^{n} {}^{n}C_r x^r\right)$ $n(1+x)^{n-1} = \sum_{r=0}^{n} r \cdot {}^{n}C_r x^{r-1}$ Now, substitute $x=1$ into this differentiated equation: $n(1+1)^{n-1} = \sum_{r=0}^{n} r \cdot {}^{n}C_r (1)^{r-1}$ $n \cdot 2^{n-1} = \sum_{r=0}^{n} r \cdot {}^{n}C_r$
   • Alternative Derivation (Algebraic): We know that for $r \neq 0$, $r \cdot {}^{n}C_r = r \cdot \frac{n}{r} {}^{n-1}C_{r-1} = n \cdot {}^{n-1}C_{r-1}$. So, $\sum_{r=0}^{n} r \cdot {}^{n}C_r = 0 \cdot {}^{n}C_0 + \sum_{r=1}^{n} r \cdot {}^{n}C_r$ $= \sum_{r=1}^{n} n \cdot {}^{n-1}C_{r-1}$ $= n \sum_{r=1}^{n} {}^{n-1}C_{r-1}$ Let $k = r-1$. When $r=1, k=0$. When $r=n, k=n-1$. $= n \sum_{k=0}^{n-1} {}^{n-1}C_k = n \cdot 2^{n-1}$ (using the first identity for $n-1$).

Understanding the Series and Finding the General Term The given series is: $S = 2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$

First, we need to express this series in a compact summation form by identifying its general term. Each term in the series is of the form $(\text{coefficient}) \cdot {}^{20}C_r$. The binomial coefficients are ${}^{20}C_0, {}^{20}C_1, {}^{20}C_2, \dots, {}^{20}C_{20}$, which means the index $r$ ranges from $0$ to $20$.

Now let's examine the coefficients multiplying these binomial terms: For $r=0$, the coefficient is $2$. For $r=1$, the coefficient is $5$. For $r=2$, the coefficient is $8$. For $r=3$, the coefficient is $11$.

This sequence of coefficients ($2, 5, 8, 11, \dots$) is an arithmetic progression (A.P.) because the difference between consecutive terms is constant: $5-2=3$, $8-5=3$, $11-8=3$. The first term of this A.P. is $a = 2$, and the common difference is $d = 3$. The $r$-th term (when starting counting from $r=0$) of an A.P. is given by $a + (r)d$. So, the coefficient of ${}^{20}C_r$ is $2 + (r)3 = 3r + 2$.

Therefore, the general term of the series, $T_r$, is $(3r+2){}^{20}C_r$. The sum of the series, $S$, can be written as: $S = \sum_{r=0}^{20} (3r+2) {}^{20}C_r$

Deconstructing the Summation To apply our known binomial identities, we can split this summation into two separate sums using the linearity property of summation ($\sum (A_r + B_r) = \sum A_r + \sum B_r$ and $\sum c \cdot A_r = c \sum A_r$): $S = \sum_{r=0}^{20} (3r \cdot {}^{20}C_r + 2 \cdot {}^{20}C_r)$ $S = \sum_{r=0}^{20} 3r \cdot {}^{20}C_r + \sum_{r=0}^{20} 2 \cdot {}^{20}C_r$ We can pull the constant factors out of the summation: $S = 3 \sum_{r=0}^{20} r \cdot {}^{20}C_r + 2 \sum_{r=0}^{20} {}^{20}C_r$ This step is crucial as it transforms the original complex sum into two standard forms for which we have direct identities.

Applying Binomial Identities Now we apply the identities discussed in the "Key Concepts and Formulas" section. In this problem, $n=20$.

1. For the second sum, $\sum\limits_{r=0}^{20} {}^{20}C_r$: Using the identity $\sum\limits_{r=0}^{n} {}^{n}C_r = 2^n$, with $n=20$: $\sum_{r=0}^{20} {}^{20}C_r = 2^{20}$

2. For the first sum, $\sum\limits_{r=0}^{20} r \cdot {}^{20}C_r$: Using the identity $\sum\limits_{r=0}^{n} r \cdot {}^{n}C_r = n \cdot 2^{n-1}$, with $n=20$: $\sum_{r=0}^{20} r \cdot {}^{20}C_r = 20 \cdot 2^{20-1} = 20 \cdot 2^{19}$

Substitute these results back into the expression for $S$: $S = 3 \cdot (20 \cdot 2^{19}) + 2 \cdot (2^{20})$

Simplification and Calculation Now, we perform the arithmetic and simplify the expression for $S$: $S = 60 \cdot 2^{19} + 2 \cdot 2^{20}$ To combine these terms, it's beneficial to express both parts with the same power of $2$. We can rewrite $2 \cdot 2^{20}$ as $2^{1} \cdot 2^{20} = 2^{1+20} = 2^{21}$. So the expression becomes: $S = 60 \cdot 2^{19} + 2^{21}$ Now, we want to factor out a common power of $2$, ideally the highest common power, which is $2^{19}$. Or, to align with the provided solution, we can try to factor out $2^{21}$. To do this, rewrite $60 \cdot 2^{19}$ in terms of $2^{21}$. We know that $2^{19} = \frac{2^{21}}{2^2} = \frac{2^{21}}{4}$. Substitute this into the equation: $S = 60 \cdot \left(\frac{2^{21}}{4}\right) + 2^{21}$ $S = 15 \cdot 2^{21} + 2^{21}$ Now, factor out $2^{21}$: $S = 2^{21} (15 + 1)$ $S = 2^{21} \cdot 16$ Since $16$ can be expressed as a power of $2$, specifically $16 = 2^4$: $S = 2^{21} \cdot 2^4$ Using the exponent rule $a^m \cdot a^n = a^{m+n}$: $S = 2^{21+4}$ $S = 2^{25}$

Tips and Common Mistakes to Avoid

• Incorrect General Term: Always double-check your arithmetic progression formula to ensure the general term correctly represents the coefficients for all $r$ values from $0$ to $n$.
• Misapplication of Identities: Ensure you use the correct binomial identity for each part of the sum. Forgetting the $n$ or $n-1$ in the $n \cdot 2^{n-1}$ formula is a common error.
• Errors in Exponent Manipulation: Be very careful when combining terms with different powers of $2$. For instance, $2 \cdot 2^{20}$ is $2^{21}$, not $4^{20}$ or $2^{40}$. When factoring, ensure you correctly divide the coefficient by the factored power of $2$ (e.g., $60 \cdot 2^{19} = (15 \cdot 4) \cdot 2^{19} = 15 \cdot 2^2 \cdot 2^{19} = 15 \cdot 2^{21}$).
• Forgetting $r=0$ term: The identity $\sum r \cdot {}^{n}C_r$ correctly accounts for the $r=0$ term (which is $0 \cdot {}^{n}C_0 = 0$), so you don't need to explicitly exclude it from the summation.

Summary/Key Takeaway This problem is a classic example of evaluating sums of series involving binomial coefficients multiplied by terms forming an arithmetic progression. The primary strategy involves:

1. Identifying the general term of the series.
2. Decomposing the complex summation into simpler sums that match known binomial identities.
3. Applying these standard identities: $\sum {}^{n}C_r = 2^n$ and $\sum r \cdot {}^{n}C_r = n \cdot 2^{n-1}$.
4. Performing careful algebraic simplification, especially when dealing with powers of $2$, to arrive at the final answer. Mastering these identities and algebraic manipulations is crucial for solving such problems efficiently.