Find the sum to n terms of the series :`1^2+(1^2+2^2)+(1^2+2^2+3^2)+dotdotdot`

Explanation

To find the sum to n terms of the series \( S_n = 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + \ldots + (1^2 + 2^2 + 3^2 + \ldots + n^2) \), we can follow these steps:

### Step 1: Understand the Series
The series can be rewritten as:
\[ S_n = \sum_{r=1}^{n} \left( \sum_{k=1}^{r} k^2 \right) \]
where \( \sum_{k=1}^{r} k^2 \) is the sum of the squares of the first \( r \) natural numbers.

### Step 2: Use the Formula for the Sum of Squares
The formula for the sum of the squares of the first \( r \) natural numbers is:
\[ \sum_{k=1}^{r} k^2 = \frac{r(r+1)(2r+1)}{6} \]
Thus, we can express \( S_n \) as:
\[ S_n = \sum_{r=1}^{n} \frac{r(r+1)(2r+1)}{6} \]

### Step 3: Factor Out the Constant
We can factor out \( \frac{1}{6} \):
\[ S_n = \frac{1}{6} \sum_{r=1}^{n} r(r+1)(2r+1) \]

### Step 4: Expand the Expression
Now we need to compute the sum \( \sum_{r=1}^{n} r(r+1)(2r+1) \). We can expand \( r(r+1)(2r+1) \):
\[ r(r+1)(2r+1) = 2r^3 + 3r^2 + r \]
Thus,
\[ S_n = \frac{1}{6} \left( \sum_{r=1}^{n} (2r^3 + 3r^2 + r) \right) \]

### Step 5: Break Down the Summation
We can break this down into three separate sums:
\[ S_n = \frac{1}{6} \left( 2 \sum_{r=1}^{n} r^3 + 3 \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \right) \]

### Step 6: Use Known Summation Formulas
We can use the following formulas:
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. \( \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \)

Substituting these into our expression:
\[ S_n = \frac{1}{6} \left( 2 \left( \frac{n(n+1)}{2} \right)^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right) \]

### Step 7: Simplify the Expression
Calculating each term:
1. \( 2 \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{2} \)
2. \( 3 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{2} \)
3. \( \frac{n(n+1)}{2} \)

Combining these:
\[ S_n = \frac{1}{6} \left( \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right) \]
Factoring out \( \frac{n(n+1)}{2} \):
\[ S_n = \frac{n(n+1)}{12} \left( n(n+1) + (2n+1) + 1 \right) \]

### Step 8: Final Simplification
This simplifies to:
\[ S_n = \frac{n(n+1)(n^2 + 3n + 2)}{12} = \frac{n(n+1)(n+1)(n+2)}{12} = \frac{n(n+1)^2(n+2)}{12} \]

### Final Result
Thus, the sum to \( n \) terms of the series is:
\[ S_n = \frac{n(n+1)(n+2)(n+1)}{12} \]