The sum to 100 terms of the series
` 1.2.3. + 2.3.4. + 3.4.5. + …+ n(n + 1) (n+2) + ….`
is integral multiple of

Options

A. 2525
B. 2526
C. 2527
D. 2528

Answer

To solve the problem, we need to find the sum of the first 100 terms of the series given by:

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

### Step 1: Identify the general term
The general term of the series can be expressed as:

\[ a_n = n(n + 1)(n + 2) \]

### Step 2: Write the sum of the first n terms
The sum of the first n terms can be expressed as:

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

### Step 3: Expand the general term
We can expand the general term:

\[ k(k + 1)(k + 2) = k(k^2 + 3k + 2) = k^3 + 3k^2 + 2k \]

### Step 4: Write the sum in terms of known formulas
Now we can write the sum as:

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

This can be separated into three sums:

\[ S_n = \sum_{k=1}^{n} k^3 + 3\sum_{k=1}^{n} k^2 + 2\sum_{k=1}^{n} k \]

### Step 5: Use formulas for the sums
We can use the formulas for the sums of powers:

1. \( \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \)
2. \( \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \)
3. \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2 \)

### Step 6: Substitute the formulas into the sum
Substituting these formulas into our expression for \( S_n \):

\[ S_n = \left( \frac{n(n + 1)}{2} \right)^2 + 3 \cdot \frac{n(n + 1)(2n + 1)}{6} + 2 \cdot \frac{n(n + 1)}{2} \]

### Step 7: Simplify the expression
Now we simplify each term:

1. The first term becomes \( \frac{n^2(n + 1)^2}{4} \)
2. The second term simplifies to \( \frac{n(n + 1)(2n + 1)}{2} \)
3. The third term simplifies to \( n(n + 1) \)

Combining these, we get:

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

### Step 8: Factor out common terms
Factoring out \( n(n + 1) \):

\[ S_n = n(n + 1) \left( \frac{n(n + 1)}{4} + \frac{2n + 1}{2} + 1 \right) \]

### Step 9: Find the value for n = 100
Now, we substitute \( n = 100 \):

\[ S_{100} = 100(101) \left( \frac{100(101)}{4} + \frac{201}{2} + 1 \right) \]

Calculating each term:

1. \( \frac{100(101)}{4} = 2525 \)
2. \( \frac{201}{2} = 100.5 \)
3. Adding these gives \( 2525 + 100.5 + 1 = 2626.5 \)

### Step 10: Final calculation
Now, we compute:

\[ S_{100} = 100 \cdot 101 \cdot 2626.5 \]

### Conclusion
Since we are interested in the integral multiple, we observe that the sum \( S_{100} \) is an integral multiple of \( 2525 \).