Key Concepts and Formulas

• Binomial Theorem: $(x+y)^n = \sum_{r=0}^n \binom{n}{r} x^{n-r} y^r$. Setting $x=1, y=1$ gives $\sum_{r=0}^n \binom{n}{r} = 2^n$.
• Properties of Binomial Coefficients:
  • $r \binom{n}{r} = n \binom{n-1}{r-1}$
  • $r(r-1) \binom{n}{r} = n(n-1) \binom{n-2}{r-2}$
• Integration of Binomial Expansion: Integrating $(1+x)^n$ term-by-term allows evaluation of sums of the form $\sum_{r=0}^n \frac{\binom{n}{r}}{r+1}$.

Step-by-Step Solution

Step 1: Evaluate $\alpha$. We are given $\alpha = \sum_{r=0}^n (4r^2 + 2r + 1) {}^n C_r$. We can split this sum into three parts: $\alpha = 4 \sum_{r=0}^n r^2 {}^n C_r + 2 \sum_{r=0}^n r {}^n C_r + \sum_{r=0}^n {}^n C_r$ We will evaluate each sum separately.

Step 1.1: Evaluate $\sum_{r=0}^n {}^n C_r$. Using the binomial theorem with $x=1, y=1$, we have: $\sum_{r=0}^n {}^n C_r = (1+1)^n = 2^n$

Step 1.2: Evaluate $\sum_{r=0}^n r {}^n C_r$. We use the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$. $\sum_{r=0}^n r {}^n C_r = \sum_{r=1}^n r \frac{n!}{r!(n-r)!} = \sum_{r=1}^n \frac{n!}{(r-1)!(n-r)!} = \sum_{r=1}^n n \frac{(n-1)!}{(r-1)!(n-r)!} = n \sum_{r=1}^n \binom{n-1}{r-1}$ Let $k = r-1$. When $r=1$, $k=0$. When $r=n$, $k=n-1$. $n \sum_{k=0}^{n-1} \binom{n-1}{k} = n 2^{n-1}$

Step 1.3: Evaluate $\sum_{r=0}^n r^2 {}^n C_r$. We use the identities $r \binom{n}{r} = n \binom{n-1}{r-1}$ and $r^2 \binom{n}{r} = r \left( n \binom{n-1}{r-1} \right) = n \left( (r-1)+1 \right) \binom{n-1}{r-1} = n(r-1) \binom{n-1}{r-1} + n \binom{n-1}{r-1}$. Using $r-1 \binom{n-1}{r-1} = (n-1) \binom{n-2}{r-2}$: $r^2 \binom{n}{r} = n (n-1) \binom{n-2}{r-2} + n \binom{n-1}{r-1}$ Now, summing over $r$: $\sum_{r=0}^n r^2 \binom{n}{r} = \sum_{r=2}^n n(n-1) \binom{n-2}{r-2} + \sum_{r=1}^n n \binom{n-1}{r-1}$ (The summation starts from $r=2$ for the first term because for $r<2$, $r(r-1)=0$. The summation starts from $r=1$ for the second term because for $r=0$, $r\binom{n}{r}=0$.) $= n(n-1) \sum_{r=2}^n \binom{n-2}{r-2} + n \sum_{r=1}^n \binom{n-1}{r-1}$ Let $k = r-2$ in the first sum and $j = r-1$ in the second sum. $= n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k} + n \sum_{j=0}^{n-1} \binom{n-1}{j}$ $= n(n-1) 2^{n-2} + n 2^{n-1}$ $= n(n-1) 2^{n-2} + 2n 2^{n-2} = (n^2 - n + 2n) 2^{n-2} = (n^2+n) 2^{n-2} = n(n+1) 2^{n-2}$

Step 1.4: Combine the sums to find $\alpha$. $\alpha = 4 \left( n(n+1) 2^{n-2} \right) + 2 \left( n 2^{n-1} \right) + 2^n$ $\alpha = n(n+1) 2^n + n 2^n + 2^n$ $\alpha = 2^n (n(n+1) + n + 1) = 2^n (n^2 + n + n + 1) = 2^n (n^2 + 2n + 1) = 2^n (n+1)^2$

Step 2: Evaluate $\beta$. We are given $\beta = \sum_{r=0}^n \frac{{}^n C_r}{r+1} + \frac{1}{n+1}$. Consider the binomial expansion: $(1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r$. Integrate both sides with respect to $x$ from 0 to 1: $\int_0^1 (1+x)^n dx = \int_0^1 \left( \sum_{r=0}^n \binom{n}{r} x^r \right) dx$ $\left[ \frac{(1+x)^{n+1}}{n+1} \right]_0^1 = \sum_{r=0}^n \binom{n}{r} \int_0^1 x^r dx$ $\frac{(1+1)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \sum_{r=0}^n \binom{n}{r} \left[ \frac{x^{r+1}}{r+1} \right]_0^1$ $\frac{2^{n+1} - 1}{n+1} = \sum_{r=0}^n \frac{\binom{n}{r}}{r+1}$ Now substitute this back into the expression for $\beta$: $\beta = \frac{2^{n+1} - 1}{n+1} + \frac{1}{n+1} = \frac{2^{n+1}}{n+1}$

Step 3: Evaluate $\frac{2\alpha}{\beta}$. Substitute the expressions for $\alpha$ and $\beta$: $\frac{2\alpha}{\beta} = \frac{2 \cdot 2^n (n+1)^2}{\frac{2^{n+1}}{n+1}} = \frac{2^{n+1} (n+1)^2}{\frac{2^{n+1}}{n+1}}$ $\frac{2\alpha}{\beta} = (n+1)^2 \cdot (n+1) = (n+1)^3$

Step 4: Solve the inequality for $n$. We are given $140 < \frac{2\alpha}{\beta} < 281$. Substituting our result from Step 3: $140 < (n+1)^3 < 281$ To find $n$, we take the cube root of all parts of the inequality: $\sqrt[3]{140} < n+1 < \sqrt[3]{281}$ We know that $5^3 = 125$ and $6^3 = 216$, and $7^3 = 343$. So, $\sqrt[3]{140}$ is slightly larger than 5, and $\sqrt[3]{281}$ is between 6 and 7. More precisely, $\sqrt[3]{140} \approx 5.19$ and $\sqrt[3]{281} \approx 6.55$. Thus, we have: $5.19 < n+1 < 6.55$ Since $n$ must be a positive integer, $n+1$ must be an integer. The only integer between 5.19 and 6.55 is 6. So, $n+1 = 6$. Solving for $n$: $n = 6 - 1 = 5$

Common Mistakes & Tips

• Algebraic Errors: Be very careful with algebraic manipulations, especially when dealing with factorials and exponents.
• Summation Limits: Pay close attention to the starting and ending indices of summations when using identities like $r \binom{n}{r} = n \binom{n-1}{r-1}$, as they can affect the summation limits.
• Integration Constant: When performing indefinite integration, the constant of integration is not needed when evaluating definite integrals between limits.

Summary

The problem required evaluating two complex sums, $\alpha$ and $\beta$, involving binomial coefficients. We successfully simplified $\alpha$ by breaking down the summand and using standard identities for sums of $r \binom{n}{r}$ and $r^2 \binom{n}{r}$. The expression for $\beta$ was simplified by integrating the binomial expansion of $(1+x)^n$. After finding expressions for $\alpha$ and $\beta$, we computed $\frac{2\alpha}{\beta}$ and used the given inequality to solve for the integer value of $n$. The solution involved recognizing that $(n+1)^3$ must be an integer whose cube root falls within a specific range, leading to a unique integer value for $n$.

The final answer is $\boxed{5}$.