The sum to n terms of the series
`(n^(2)-1^(2))+2(n^(2)-2^(2))+3(n^(2)-3^(2))+ . . . .`, is

Options

A. `(n^(2))/(4)(n^(2)-1)`
B. `(n)/(4)(n+1)^(2)`
C. 0
D. `2n(n^(2)-1)`

Answer

To find the sum to \( n \) terms of the series

\[ (n^2 - 1^2) + 2(n^2 - 2^2) + 3(n^2 - 3^2) + \ldots + n(n^2 - n^2), \]

we can start by rewriting the general term of the series.

### Step 1: Rewrite the General Term
The general term of the series can be expressed as:

\[ k(n^2 - k^2) \quad \text{for } k = 1, 2, \ldots, n. \]

This can be simplified to:

\[ k(n^2 - k^2) = k(n^2 - k^2) = kn^2 - k^3. \]

### Step 2: Sum the Series
Now, we can express the sum \( S_n \) of the first \( n \) terms as follows:

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

### Step 3: Use Known Formulas
We know the formulas for the sums:

1. The sum of the first \( n \) natural numbers:

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

2. 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 4: Substitute the Formulas
Substituting these formulas into our expression for \( S_n \):

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

### Step 5: Simplify the Expression
Now, we simplify \( S_n \):

1. The first term:

\[ n^2 \cdot \frac{n(n+1)}{2} = \frac{n^3(n+1)}{2}. \]

2. The second term:

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

Now, substituting these back into \( S_n \):

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

### Step 6: Find a Common Denominator
To combine these fractions, we find a common denominator:

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

### Step 7: Factor the Numerator
Factoring out \( n^2(n+1) \):

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

### Final Result
Thus, the sum to \( n \) terms of the series is:

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