Animated Solution for Mathematics - Binomial Theorem: Let ${}^n{C}_r$ denote the binomial coefficient of $x^r$ in the expansion of $(1+x{)}^n$. If ${\sum }_{k=0}^{10}(2^2+3k){}^{10}{C}_k=\alpha \cdot 3^{10}+\beta \cdot 2^{10},\alpha ,\beta \in \mathbb{R}$, then $\alpha +\beta$ is equal to ___

Welcome, aspiring mathematician! Today, we are going to peel back the layers of a classic binomial summation problem. These problems often look like a tangled mess of variables and indices, but they are actually elegant puzzles waiting to be solved.

Let us look at the expression we are tackling:

The Art of Simplification

Before we dive into the deep end, let us clean up our workspace. The term $2^2$ is just a constant, $4$. So, our summation becomes ${\sum }_{k=0}^{10}(4+3k){}^{10}{C}_k$.

Now, imagine you are standing before a large, complex task. The best strategy is to break it down. We use the linearity of summation to split this into two manageable pieces:

By distributing the summation, we have turned one intimidating problem into two standard, well-known identities. This is the secret to success in competitive exams—recognizing the patterns you already know.

The Toolkit of Identities

Now, we reach into our mathematical toolkit. We need two essential identities for binomial coefficients.

The first is the sum of all coefficients:

The second, which accounts for the index $k$, is:

With $n=10$, these become our guiding stars.

For the first part, $4{\sum }_{k=0}^{10}{}^{10}{C}_k$, we simply substitute $n=10$ to get $4\cdot 2^{10}$.

For the second part, $3{\sum }_{k=0}^{10}k\cdot {}^{10}{C}_k$, we use our second identity to get $3\cdot (10\cdot 2^{10-1})$, which simplifies to $3\cdot 10\cdot 2^9$, or $30\cdot 2^9$.

The Exponent Matching Trick

We are almost there! We have $4\cdot 2^{10}+30\cdot 2^9$. To combine these, we need the exponents to match.

This is a common trap where students get stuck. We can rewrite $30\cdot 2^9$ as $(15\cdot 2)\cdot 2^9$, which is $15\cdot 2^{10}$.

Now, our expression is $4\cdot 2^{10}+15\cdot 2^{10}$. Adding these is now trivial:

The Final Comparison

We compare our result, $19\cdot 2^{10}$, with the given form $\alpha \cdot 3^{10}+\beta \cdot 2^{10}$. It is clear that $\alpha =0$ and $\beta =19$.

The sum $\alpha +\beta$ is simply $0+19=19$.

You have successfully navigated the complexity and arrived at the truth. Keep practicing this systematic approach, and you will find that even the most daunting problems become simple, logical steps. The final answer is 19.