Question

If $n \geq 2$ is a positive integer, then the sum of the series ${ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots .+{ }^{n} C_{2}\right)$ is:

• A. $\frac{ n ( n -1)(2 n +1)}{6}$
• B. $\frac{ n ( n +1)(2 n +1)}{6}$
• C. $\frac{ n (2 n +1)(3 n +1)}{6}$
• D. $\frac{ n ( n +1)^{2}( n +2)}{12}$

Answer 1

The Correct Option is B

Step-by-Step Solution:

We are given the series ${ }^{n+1} C_{2} + 2 \left({ }^{2} C_{2} + { }^{3} C_{2} + { }^{4} C_{2} + \ldots + { }^{n} C_{2}\right)$, and we need to find its sum.

Breaking Down the Terms:

• {}{n+1} C_2 represents the number of ways to choose 2 objects from \( n+1 \) objects.
• Generally, {}^r C_2 = \frac{r(r-1)}{2}. So:
  • {}^{n+1} C_2 = \frac{(n+1)n}{2}.
  • Now, calculate the series {}{2} C_{2} + {}^{3} C_{2} + \cdots + {}^{n} C_{2}.

We can express each term {}^r C_2 as \frac{r(r-1)}{2}. Thus the series becomes:

• \frac{2 \cdot 1}{2} + \frac{3 \cdot 2}{2} + \cdots + \frac{n(n-1)}{2}
• This simplifies to 1 + 3 + 6 + \cdots + \frac{n(n-1)}{2}.

Finding the Sum of the Series:

The term \frac{r(r-1)}{2} for r = 2 \text{ to } n translates to finding the sum:

• \sum_{r=2}^{n} \frac{r(r-1)}{2} = \sum_{r=2}^{n} \frac{r^2 - r}{2}

Using the known formula for the sum of first \(m\) natural numbers square and sum:

• \sum_{r=1}^{m} r = \frac{m(m+1)}{2}
• \sum_{r=1}^{m} r^2 = \frac{m(m+1)(2m+1)}{6}

We calculate:

• \sum_{r=2}^{n} r^2 = \sum_{r=1}^{n} r^2 - 1^2 = \frac{n(n+1)(2n+1)}{6} - 1
• \sum_{r=2}^{n} r = \sum_{r=1}^{n} r - 1 = \frac{n(n+1)}{2} - 1

Putting it together:

\frac{1}{2} \left( \sum_{r=2}^{n} r^2 - \sum_{r=2}^{n} r \right) = \frac{1}{2} \left(\frac{n(n+1)(2n+1)}{6} - 1 - \frac{n(n+1)}{2} + 1 \right)

Complete Series Sum:

Finally, combining it all:

• {}^{n+1} C_2 + 2 \cdot \left( \sum_{r=2}^{n} \frac{r(r-1)}{2} \right)
• Substitute the previously found components and simplify, yielding the sum as:

\frac{n(n+1)(2n+1)}{6}.

Conclusion:

Thus, the correct answer is: \frac{ n ( n +1)(2 n +1)}{6}