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

Answer

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 will follow these steps:

### Step 1: Identify the nth term of the series
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 S_n
The total sum \( S_n \) can be expressed as:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n \]
Substituting the formula for \( T_k \) into the sum gives:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k(k + 1)(2k + 1)}{6} \]

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

### Step 5: Expand the summation
Now we need to compute the sum \( \sum_{k=1}^{n} k(k + 1)(2k + 1) \). We can expand \( k(k + 1)(2k + 1) \):
\[ k(k + 1)(2k + 1) = 2k^3 + 3k^2 + k \]
Thus,
\[ S_n = \frac{1}{6} \left( 2 \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right) \]

### Step 6: Use the formulas for the sums
Using the formulas for the sums:
- \( \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \)
- \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2 \)

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

### Step 7: Simplify the expression
Now we simplify:
1. The first term becomes \( \frac{n^2(n + 1)^2}{6} \)
2. The second term simplifies to \( \frac{n(n + 1)(2n + 1)}{4} \)
3. The third term simplifies to \( \frac{n(n + 1)}{12} \)

Combining these terms under a common denominator of 12:
\[ S_n = \frac{1}{6} \left( \frac{2n^2(n + 1)^2 + 3n(n + 1)(2n + 1) + n(n + 1)}{12} \right) \]

### Step 8: Final expression
After simplifying, we find:
\[ S_n = \frac{n(n + 1)(n + 2)(n + 1)}{12} \]

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

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