Find the sum `1^2+(1^2+2^2)+(1^2+2^2+3^2)+` up to 22nd term.

Explanation

To find the sum of the series \( S = 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + \ldots \) up to the 22nd term, we can break down the problem step by step.

### Step 1: Identify the nth term
The nth term of the series can be expressed as:
\[ T_n = 1^2 + 2^2 + 3^2 + \ldots + n^2 \]
This is the sum of the squares of the first n natural numbers.

### Step 2: Use the formula for the sum of squares
The formula for the sum of the squares of the first n natural numbers is:
\[ T_n = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 3: Write the total sum up to the 22nd term
We want to find:
\[ S = T_1 + T_2 + T_3 + \ldots + T_{22} \]
This can be rewritten as:
\[ S = \sum_{n=1}^{22} T_n = \sum_{n=1}^{22} \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Factor out the constant
We can factor out \(\frac{1}{6}\):
\[ S = \frac{1}{6} \sum_{n=1}^{22} n(n + 1)(2n + 1) \]

### Step 5: Expand the summation
Now, we need to expand \(n(n + 1)(2n + 1)\):
\[ n(n + 1)(2n + 1) = 2n^3 + 3n^2 + n \]
Thus, we can write:
\[ S = \frac{1}{6} \left( \sum_{n=1}^{22} (2n^3 + 3n^2 + n) \right) \]

### Step 6: Break down the summation
Now, we can separate the summation:
\[ S = \frac{1}{6} \left( 2 \sum_{n=1}^{22} n^3 + 3 \sum_{n=1}^{22} n^2 + \sum_{n=1}^{22} n \right) \]

### Step 7: Use summation formulas
We will use the following formulas:
- The sum of the first n natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- The sum of the squares of the first n natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- The sum of the cubes of the first n natural numbers:
\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2 \]

### Step 8: Calculate each summation for n = 22
1. **Sum of cubes**:
\[ \sum_{n=1}^{22} n^3 = \left( \frac{22 \cdot 23}{2} \right)^2 = 253^2 = 64009 \]
2. **Sum of squares**:
\[ \sum_{n=1}^{22} n^2 = \frac{22 \cdot 23 \cdot 45}{6} = 2555 \]
3. **Sum of natural numbers**:
\[ \sum_{n=1}^{22} n = \frac{22 \cdot 23}{2} = 253 \]

### Step 9: Substitute back into the equation
Now substituting these values back into our equation for S:
\[ S = \frac{1}{6} \left( 2 \cdot 64009 + 3 \cdot 2555 + 253 \right) \]
Calculating this gives:
\[ S = \frac{1}{6} \left( 128018 + 7665 + 253 \right) = \frac{1}{6} \left( 135936 \right) = 22656 \]

### Final Answer
Thus, the sum up to the 22nd term is:
\[ \boxed{22656} \]