if n>=2, `""^(n+1)C_(2)+2(""^(2)C_(2)+""^(3)C_(2)+""^(4)C_(2)+....+""^(n)C_(2))`Value is:-

Options

A. `sumn`
B. `sumn^(2)`
C. `sumn^(3)`
D. `((n+1))/(2)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \binom{n+1}{2} + 2\left(\binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \ldots + \binom{n}{2}\right) \]

### Step 1: Evaluate \(\binom{n+1}{2}\)

The binomial coefficient \(\binom{n+1}{2}\) can be calculated using the formula:

\[ \binom{n+1}{2} = \frac{(n+1)n}{2} \]

### Step 2: Evaluate the summation

Next, we need to evaluate the summation:

\[ \sum_{r=2}^{n} \binom{r}{2} \]

Using the formula for the binomial coefficient, we have:

\[ \binom{r}{2} = \frac{r(r-1)}{2} \]

Thus, the summation becomes:

\[ \sum_{r=2}^{n} \binom{r}{2} = \sum_{r=2}^{n} \frac{r(r-1)}{2} = \frac{1}{2} \sum_{r=2}^{n} r(r-1) \]

### Step 3: Simplifying the summation

The term \(r(r-1)\) can be expressed as:

\[ r(r-1) = r^2 - r \]

So we can split the summation:

\[ \sum_{r=2}^{n} r(r-1) = \sum_{r=2}^{n} r^2 - \sum_{r=2}^{n} r \]

Using the formulas for the sums of squares and the sums of integers:

1. The sum of the first \(n\) squares is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

2. The sum of the first \(n\) integers is:

\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]

Thus, we can calculate:

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

And:

\[ \sum_{r=2}^{n} r = \sum_{r=1}^{n} r - 1 = \frac{n(n+1)}{2} - 1 \]

### Step 4: Combine the results

Now substituting back into our expression:

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

This simplifies to:

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

### Step 5: Final expression

Now, substituting this back into our original expression:

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

After simplification, we will arrive at:

\[ \text{Final Value} = \frac{n(n+1)(2n+1)}{6} \]