The value of
` 1xx2xx3xx4+2xx3xx4xx5+3xx4xx5xx6+…+n(n +1) (n +2) (n +3)`, is

Options

A. `(1)/(5) (n+1) (n+2) (n+3) (n+4) (n+5)`
B. `(1)/(5) n(n +1) (n+2) (n+3) (n+4)`
C. `(1)/(5) n(n +1) (n+2) (n+3) (n+4)`
D. `""^(n+4)C_(5)`

Answer

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

\[ S_n = 1 \times 2 \times 3 \times 4 + 2 \times 3 \times 4 \times 5 + 3 \times 4 \times 5 \times 6 + \ldots + n(n + 1)(n + 2)(n + 3) \]

### Step-by-Step Solution:

**Step 1: Rewrite the terms in factorial form.**

Each term in the series can be expressed in terms of factorials. For example:
- The first term \(1 \times 2 \times 3 \times 4\) can be written as \(4!\).
- The second term \(2 \times 3 \times 4 \times 5\) can be written as \(\frac{5!}{1!}\).
- The third term \(3 \times 4 \times 5 \times 6\) can be written as \(\frac{6!}{2!}\).
- The general term \(k(k + 1)(k + 2)(k + 3)\) can be expressed as \(\frac{(k + 3)!}{(k - 1)!}\).

Thus, we can rewrite \(S_n\) as:

\[ S_n = \sum_{k=1}^{n} \frac{(k + 3)!}{(k - 1)!} \]

**Step 2: Factor out common terms.**

Notice that we can factor out \(4!\) from each term:

\[ S_n = 4! \sum_{k=1}^{n} \frac{(k + 3)!}{4! \cdot (k - 1)!} \]

This simplifies to:

\[ S_n = 4! \sum_{k=1}^{n} \frac{(k + 3)(k + 2)(k + 1)}{3!} \]

**Step 3: Change the index of summation.**

Let \(j = k + 3\), then \(k = j - 3\). When \(k = 1\), \(j = 4\) and when \(k = n\), \(j = n + 3\). Thus, we can rewrite the sum:

\[ S_n = 4! \sum_{j=4}^{n + 3} \frac{j(j - 1)(j - 2)}{3!} \]

**Step 4: Simplify the sum.**

The expression \(\frac{j(j - 1)(j - 2)}{3!}\) counts the number of ways to choose 3 objects from \(j\), which is \(C(j, 3)\). Therefore, we can rewrite the sum as:

\[ S_n = 4! \sum_{j=4}^{n + 3} C(j, 3) \]

**Step 5: Use the hockey-stick identity.**

Using the hockey-stick identity in combinatorics, we have:

\[ \sum_{j=r}^{n} C(j, r) = C(n + 1, r + 1) \]

Thus, we can write:

\[ \sum_{j=4}^{n + 3} C(j, 3) = C(n + 4, 4) \]

**Step 6: Final expression for \(S_n\).**

Substituting this back into our expression for \(S_n\):

\[ S_n = 4! \cdot C(n + 4, 4) \]

Calculating \(4!\):

\[ 4! = 24 \]

Thus, we have:

\[ S_n = 24 \cdot \frac{(n + 4)(n + 3)(n + 2)(n + 1)}{4!} \]

This simplifies to:

\[ S_n = \frac{(n + 4)(n + 3)(n + 2)(n + 1)}{5} \]

### Final Answer:

\[ S_n = \frac{n(n + 1)(n + 2)(n + 3)}{5} \]