Sum up to n terms the series `1.2^(2)+2.3^(2)+ 3.4^(2)+...`

Answer

To find the sum of the series \( S_n = 1 \cdot 2^2 + 2 \cdot 3^2 + 3 \cdot 4^2 + \ldots + n \cdot (n+1)^2 \), we can express the \( r \)-th term of the series as:

\[ T_r = r \cdot (r + 1)^2 \]

### Step 1: Expand the \( r \)-th term
Expanding \( T_r \):

\[ T_r = r \cdot (r^2 + 2r + 1) = r^3 + 2r^2 + r \]

### Step 2: Write the sum of the first \( n \) terms
Now, we need to find the sum \( S_n \):

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

### Step 3: Break down the sum
We can separate the sums:

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

### Step 4: Use known formulas for sums
We use the following formulas for the sums:

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

### Step 5: Substitute the formulas into the sum
Substituting these formulas into \( S_n \):

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

### Step 6: Simplify each term
1. The first term is:

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

2. The second term simplifies to:

\[ \frac{2n(n + 1)(2n + 1)}{6} = \frac{n(n + 1)(2n + 1)}{3} \]

3. The third term is:

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

### Step 7: Combine all terms
Now we combine all three terms:

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

### Step 8: Find a common denominator
The common denominator for these fractions is \( 12 \):

\[ S_n = \frac{3n^2(n + 1)^2}{12} + \frac{4n(n + 1)(2n + 1)}{12} + \frac{6n(n + 1)}{12} \]

### Step 9: Combine the numerators
Combining the numerators:

\[ S_n = \frac{3n^2(n + 1)^2 + 4n(n + 1)(2n + 1) + 6n(n + 1)}{12} \]

### Step 10: Factor out common terms
Factoring out \( n(n + 1) \):

\[ S_n = \frac{n(n + 1)}{12} \left( 3n(n + 1) + 4(2n + 1) + 6 \right) \]

### Step 11: Simplify the expression
Simplifying the expression inside the parentheses:

\[ 3n(n + 1) + 8n + 4 + 6 = 3n^2 + 3n + 8n + 10 = 3n^2 + 11n + 10 \]

### Final Expression
Thus, the final sum \( S_n \) is:

\[ S_n = \frac{n(n + 1)(3n^2 + 11n + 10)}{12} \]