Elaborate Solution for Series Summation

This problem involves evaluating two infinite series, 'a' and 'b', and then combining their values to find the final expression. The key concepts utilized are the Maclaurin series expansion for $e^x$, properties of binomial coefficients, and techniques for summing series involving factorials, including telescoping sums.

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1. Key Concepts and Formulas

• Maclaurin Series for $e^x$: The exponential function $e^x$ can be expressed as an infinite series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$ A special case for $x=1$ gives $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots$
• Binomial Theorem Identity: The sum of binomial coefficients for a given 'n' is $2^n$: $\sum_{k=0}^n {}^n C_k = {}^n C_0 + {}^n C_1 + \ldots + {}^n C_n = 2^n$
• Binomial Coefficient Definition: ${}^n C_r = \frac{n!}{r!(n-r)!}$. Specifically, ${}^r C_2 = \frac{r(r-1)}{2}$.
• Series Manipulation: Techniques to simplify terms within a series to identify known series or create telescoping sums.

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2. Evaluate Series 'b'

The series 'b' is given by: $b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{ }^2 C_2}{2!}+\frac{{ }^3 C_0+{ }^3 C_1+{ }^3 C_2+{ }^3 C_3}{3 !}+....$

Step 2.1: Simplify the Numerators Let's examine the numerator of each fractional term:

• For the first term ($1!$ in denominator): ${}^1 C_0 + {}^1 C_1 = 1 + 1 = 2$. This is $2^1$.
• For the second term ($2!$ in denominator): ${}^2 C_0 + {}^2 C_1 + {}^2 C_2 = 1 + 2 + 1 = 4$. This is $2^2$.
• For the third term ($3!$ in denominator): ${}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3 = 1 + 3 + 3 + 1 = 8$. This is $2^3$.

Explanation: We recognize that the sum of binomial coefficients in the numerator of the $n$-th term (where $n$ corresponds to the factorial in the denominator) is equal to $2^n$ as per the Binomial Theorem identity.

Step 2.2: Rewrite Series 'b' Substituting these simplified numerators back into the series for 'b': $b = 1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \ldots$

Step 2.3: Identify Known Series This series exactly matches the Maclaurin series expansion for $e^x$ when $x=2$. Comparing with $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$, we can see that $x=2$.

Conclusion for 'b': Therefore, $b = e^2$

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3. Evaluate Series 'a'

The series 'a' is given by: $a=1+\frac{{ }^2 C_2}{3 !}+\frac{{ }^3 C_2}{4 !}+\frac{{ }^4 C_2}{5 !}+....$

Step 3.1: Express the General Term The series can be written in summation notation, starting from $r=2$. The general term is $\frac{{}^r C_2}{(r+1)!}$. First, let's expand the binomial coefficient ${}^r C_2$: ${}^r C_2 = \frac{r(r-1)}{2}$ So, the general term becomes: $\frac{r(r-1)}{2(r+1)!}$ Now, rewrite series 'a': $a = 1 + \sum_{r=2}^{\infty} \frac{r(r-1)}{2(r+1)!}$

Step 3.2: Manipulate the General Term for Simplification We aim to simplify the term $\frac{r(r-1)}{(r+1)!}$. We know that $(r+1)! = (r+1)r(r-1)(r-2)!$. A common strategy for such series is to express the numerator in terms of factors that can cancel with parts of the factorial in the denominator. We can use the identity $r(r-1) = (r+1)r - 2r$. This allows us to split the fraction into terms that are easier to sum. $\frac{r(r-1)}{(r+1)!} = \frac{(r+1)r - 2r}{(r+1)!} = \frac{(r+1)r}{(r+1)!} - \frac{2r}{(r+1)!}$ Now simplify each part:

• $\frac{(r+1)r}{(r+1)!} = \frac{(r+1)r}{(r+1)r(r-1)!} = \frac{1}{(r-1)!}$
• $\frac{2r}{(r+1)!}$ So, the general term becomes: $a = 1 + \frac{1}{2} \sum_{r=2}^{\infty} \left( \frac{1}{(r-1)!} - \frac{2r}{(r+1)!} \right)$ $a = 1 + \frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-1)!} - \sum_{r=2}^{\infty} \frac{r}{(r+1)!}$

Step 3.3: Evaluate the First Summation Consider the first sum: $\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-1)!}$ Let $k = r-1$. When $r=2$, $k=1$. As $r \to \infty$, $k \to \infty$. So the sum becomes: $\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k!} = \frac{1}{2} \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots \right)$ Explanation: This is the Maclaurin series for $e^1$ but missing the $k=0$ term ($\frac{1}{0!} = 1$). We know $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \ldots$, so $\frac{1}{1!} + \frac{1}{2!} + \ldots = e-1$. Thus, the first sum evaluates to: $\frac{1}{2}(e-1)$

Step 3.4: Evaluate the Second Summation Consider the second sum: $-\sum_{r=2}^{\infty} \frac{r}{(r+1)!}$ We can manipulate the term $\frac{r}{(r+1)!}$ further. We want to create a telescoping sum. Explanation: By writing $r$ as $(r+1)-1$, we can split the fraction. $\frac{r}{(r+1)!} = \frac{(r+1)-1}{(r+1)!} = \frac{r+1}{(r+1)!} - \frac{1}{(r+1)!} = \frac{1}{r!} - \frac{1}{(r+1)!}$ Now, the second summation becomes: $-\sum_{r=2}^{\infty} \left( \frac{1}{r!} - \frac{1}{(r+1)!} \right)$ Let's write out the first few terms to observe the telescoping nature: For $r=2: -\left(\frac{1}{2!} - \frac{1}{3!}\right)$ For $r=3: -\left(\frac{1}{3!} - \frac{1}{4!}\right)$ For $r=4: -\left(\frac{1}{4!} - \frac{1}{5!}\right)$ ... Summing these terms, intermediate terms cancel out: $-\left[ \left(\frac{1}{2!} - \frac{1}{3!}\right) + \left(\frac{1}{3!} - \frac{1}{4!}\right) + \left(\frac{1}{4!} - \frac{1}{5!}\right) + \ldots \right]$ The sum simplifies to the first part of the first term, as the $-\frac{1}{(r+1)!}$ term approaches 0 as $r \to \infty$. $-\left( \frac{1}{2!} \right) = -\frac{1}{2}$

Step 3.5: Combine Results for 'a' Now, substitute the values of the two summations back into the expression for 'a': $a = 1 + \frac{1}{2}(e-1) - \frac{1}{2}$ $a = 1 + \frac{e}{2} - \frac{1}{2} - \frac{1}{2}$ $a = 1 + \frac{e}{2} - 1$ Conclusion for 'a': $a = \frac{e}{2}$

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4. Calculate the Final Expression $\frac{2b}{a^2}$

Now we have the values for 'a' and 'b': $b = e^2$ $a = \frac{e}{2}$

Substitute these into the target expression: $\frac{2b}{a^2} = \frac{2(e^2)}{\left(\frac{e}{2}\right)^2}$ $= \frac{2e^2}{\frac{e^2}{4}}$ $= 2e^2 \times \frac{4}{e^2}$ $= 8$

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5. Tips for Success / Common Mistakes to Avoid

• Recognize Standard Series: Always look for opportunities to relate complex series to known Maclaurin series (like $e^x$, $\sin x$, $\cos x$, etc.).
• Index Alignment: Be careful with the starting index of summations. If a series looks like $e^x$ but is missing initial terms, adjust accordingly (e.g., $\sum_{k=1}^\infty \frac{1}{k!} = e-1$).
• Factorial Manipulation: When dealing with terms like $\frac{N}{(D)!}$, try to express $N$ in terms of factors present in $D$ to simplify the fraction (e.g., $r(r-1) = (r+1)r - 2r$ or $r = (r+1)-1$).
• Telescoping Sums: Identify patterns where intermediate terms cancel out. This usually happens when the general term can be written as a difference of consecutive terms, $f(k) - f(k+1)$.
• Algebraic Precision: Double-check all algebraic manipulations, especially when dealing with fractions and signs. A small error can propagate and lead to an incorrect final answer.

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6. Summary

By first identifying the series for 'b' as the Maclaurin expansion of $e^2$, we found $b=e^2$. For series 'a', we systematically broke down its general term using binomial coefficient definitions and strategic algebraic manipulations to transform it into sums of standard series and a telescoping series. This allowed us to evaluate 'a' as $\frac{e}{2}$. Finally, substituting these values into the expression $\frac{2b}{a^2}$ yielded the result of 8. The problem highlights the importance of recognizing fundamental series expansions and applying clever algebraic techniques to simplify complex summations.