Key Concept: The Hockey-Stick Identity

The problem involves the sum of binomial coefficients. A fundamental identity used to sum consecutive binomial coefficients with the same lower index is the Hockey-Stick Identity (or known as Pascal's Identity in summation form). It states that: $\sum_{k=r}^n {}^k{C_r} = {}^r{C_r} + {}^{r+1}{C_r} + \dots + {}^n{C_r} = {}^{n+1}{C_{r+1}}$ This identity is incredibly useful for simplifying sums of combinations where the upper index varies, but the lower index remains constant. It can be visualized as following a "hockey stick" path on Pascal's triangle.

Analyzing the Given Series

We are asked to find the sum of the series: $S = {}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + \dots + {}^n{C_2}} \right)$ The series consists of two main parts:

1. A single term: ${}^{n + 1}{C_2}$
2. A summation part multiplied by 2: $2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + \dots + {}^n{C_2}} \right)$

Our strategy will be to first simplify the summation part using the Hockey-Stick Identity, and then combine it with the first term for the final result.

Simplifying the Summation Part

Let's focus on the sum inside the parenthesis: $\Sigma = {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + \dots + {}^n{C_2}}$ Here, we can clearly see the pattern matching the Hockey-Stick Identity: The lower index $r$ is fixed at $2$. The upper index $k$ ranges from $2$ to $n$.

Applying the Hockey-Stick Identity: $\sum_{k=2}^n {}^k{C_2} = {}^{n+1}{C_{2+1}} = {}^{n+1}{C_3}$ So, the summation part simplifies to $2 \cdot {}^{n+1}{C_3}$.

Why this step? By recognizing the pattern of summation of combinations with a fixed lower index, we can directly apply the powerful Hockey-Stick Identity to simplify a long sum into a single binomial coefficient. This is much more efficient than trying to apply Pascal's identity iteratively many times.

Combining and Final Simplification

Now substitute the simplified sum back into the original expression for $S$: $S = {}^{n + 1}{C_2} + 2 \cdot {}^{n+1}{C_3}$ Next, we expand the binomial coefficients using the definition ${}^m{C_k} = \frac{m!}{k!(m-k)!}$: ${}^{n+1}{C_2} = \frac{(n+1)!}{2!(n+1-2)!} = \frac{(n+1)!}{2!(n-1)!} = \frac{(n+1)n(n-1)!}{2 \cdot 1 \cdot (n-1)!} = \frac{n(n+1)}{2}$ ${}^{n+1}{C_3} = \frac{(n+1)!}{3!(n+1-3)!} = \frac{(n+1)!}{3!(n-2)!} = \frac{(n+1)n(n-1)(n-2)!}{3 \cdot 2 \cdot 1 \cdot (n-2)!} = \frac{n(n+1)(n-1)}{6}$

Substitute these expanded forms back into the expression for $S$: $S = \frac{n(n+1)}{2} + 2 \cdot \frac{n(n+1)(n-1)}{6}$ Simplify the second term: $S = \frac{n(n+1)}{2} + \frac{n(n+1)(n-1)}{3}$ To combine these fractions, find a common denominator, which is $6$: $S = \frac{3 \cdot n(n+1)}{6} + \frac{2 \cdot n(n+1)(n-1)}{6}$ Now, factor out the common term $\frac{n(n+1)}{6}$: $S = \frac{n(n+1)}{6} \left[ 3 + 2(n-1) \right]$ Distribute and simplify inside the brackets: $S = \frac{n(n+1)}{6} \left[ 3 + 2n - 2 \right]$ $S = \frac{n(n+1)}{6} \left[ 2n + 1 \right]$ Finally, arrange the terms: $S = \frac{n(n+1)(2n+1)}{6}$

Why these steps? After applying the identity, the problem reduces to algebraic manipulation of binomial coefficients. Expanding them into factorials and then simplifying allows us to combine the terms effectively. Factoring out common terms like $n(n+1)$ is a standard technique to simplify expressions and arrive at a compact form, making it easier to match with the given options.

Conclusion and Discrepancy Note

Based on a rigorous application of the Hockey-Stick Identity and subsequent algebraic simplification, the sum of the given series is: $\frac{n(n+1)(2n+1)}{6}$ This result matches option (B) from the provided choices.

Important Note on Discrepancy: The problem statement indicates that the "Correct Answer" is (A) ${{n(2n + 1)(3n + 1)} \over 6}$. However, our derivation consistently leads to option (B). To verify, let's substitute a small value for $n$, say $n=2$. The original series becomes: ${}^{2+1}{C_2} + 2\left( {{}^2{C_2}} \right) = {}^3{C_2} + 2(1) = 3 + 2 = 5$ Now let's evaluate our derived result (B) for $n=2$: $\frac{2(2+1)(2 \cdot 2 + 1)}{6} = \frac{2(3)(5)}{6} = \frac{30}{6} = 5$ Our derived result matches the series value for $n=2$. Now, let's evaluate option (A) for $n=2$: ${{2(2 \cdot 2 + 1)(3 \cdot 2 + 1)} \over 6} = {{2(5)(7)} \over 6} = {{70} \over 6} = \frac{35}{3}$ Since $\frac{35}{3} \ne 5$, option (A) is incorrect for $n=2$. This confirms that the designated "Correct Answer: A" in the problem statement is inconsistent with the problem itself. The correct answer, based on mathematical derivation, is (B).

Tips and Common Mistakes to Avoid

• Recognizing Identities: Always look for patterns in sums of binomial coefficients. The Hockey-Stick Identity is common when the lower index is constant. Pascal's identity (${}^n{C_r} + {}^n{C_{r+1}} = {}^{n+1}{C_{r+1}}$) is used for combining two terms with the same upper index.
• Algebraic Precision: Be careful with factorials, multiplications, and finding common denominators. A single error can lead to an incorrect final expression.
• Verification with Small $n$: As demonstrated above, plugging in a small integer value for $n$ (like $n=2$ or $n=3$) into both the original series and the derived options can quickly help identify errors in your derivation or inconsistencies in the problem/options.
• Understanding the Scope of Summation: Ensure you correctly identify the starting and ending values of the summation index.

Summary

This problem elegantly demonstrates the application of the Hockey-Stick Identity to simplify a series of binomial coefficients. By identifying the summation pattern and applying the identity, the problem reduces to a straightforward algebraic simplification, leading to a compact and exact formula for the sum of the series. Always verify your result, especially against provided answers, to ensure consistency.