Animated Solution for Mathematics - Binomial Theorem: Let $\alpha ={\sum }_{r=0}^n(4r^2+2r+1){}^n{C}_r$ and $\beta =\left({\sum }_{r=0}^n\frac{{}^n{C}_r}{r+1}\right)+\frac{1}{n+1}$. If $140<\frac{2\alpha }{\beta }<281$, then the value of $n$ is _______

Welcome, future engineer. Today, we are not just solving a problem; we are conducting a symphony of numbers. The Binomial Theorem is often treated as a dry list of formulas, but in the hands of a master, it is a tool of immense elegance.

We are faced with a problem involving $\alpha$ and $\beta$, two summations that look intimidating at first glance. But fear not. Every complex problem is just a collection of simple truths waiting to be revealed.

We start with the expression for $\alpha$:

When you see a quadratic polynomial multiplied by a binomial coefficient, your first instinct should be to break it down. We cannot sum $r^2{}^n{C}_r$ directly, but we can sum $r{}^n{C}_r$ and $r(r-1){}^n{C}_r$ using the 'falling factorial' property.

We rewrite $4r^2+2r+1$ as $4[r(r-1)+r]+2r+1$, which simplifies to $4r(r-1)+6r+1$. Now, the summation splits into three distinct, manageable parts.

Let us tackle the middle term, $\sum r{}^n{C}_r$. We know the identity $r{}^n{C}_r=n{}^{n-1}{C}_{r-1}$. When we sum this from $r=1$ to $n$, we are essentially summing all binomial coefficients of index $n-1$, which gives us $n\cdot 2^{n-1}$.

Now for the $r^2$ term. By using our decomposition $r(r-1)+r$, we invoke the identity:

The $n(n-1)$ pulls out, and we are left with a sum of binomial coefficients that equals $2^{n-2}$. When we combine these, the powers of 2 align perfectly. After substituting these back, we find that:

Now, let us turn our attention to $\beta$:

In the world of combinatorics, division is often just integration in disguise. We use the identity:

By pulling out the constant $\frac{1}{n+1}$, we are left with a sum of binomial coefficients ${\sum }_{r=0}^n{}^{n+1}{C}_{r+1}$. This sum represents all terms of the $(n+1)$-th row of Pascal's triangle, excluding the very first term, ${}^{n+1}{C}_0$.

Since the sum of the entire row is $2^{n+1}$, our sum is $2^{n+1}-1$. When we add the final term $\frac{1}{n+1}$ from the original expression, the $-1$ and $+1$ cancel out beautifully. We are left with:

We have arrived at the final act. We need to evaluate the ratio $\frac{2\alpha }{\beta }$. Substituting our simplified forms:

Look closely at the numerator: $2\cdot 2^n$ is $2^{n+1}$. This cancels perfectly with the $2^{n+1}$ in the denominator. The $(n+1)$ in the denominator of the denominator flips up to the numerator, multiplying with $(n+1{)}^2$ to give us $(n+1{)}^3$.

We are left with the inequality $140<(n+1{)}^3<281$. We seek an integer $n$ such that $(n+1{)}^3$ is between 140 and 281.

We know $5^3=125$ (too small) and $6^3=216$ (perfect!). Thus, $n+1=6$, which means $n=5$.