Key Concepts and Formulas

• General Term of a Series: Identifying the pattern in a sequence to express its $k$-th term, $T_k$.
• Alternating Series: A series where the signs of consecutive terms alternate, often represented by $(-1)^{k-1}$ or $(-1)^k$.
• Summation of Alternating Power Series: Using known formulas for sums like $\sum_{k=1}^{n} (-1)^{k-1} k^p$.
  • $\sum_{k=1}^{n} (-1)^{k-1} k = \frac{(-1)^{n-1}(2n+1)+1}{4}$
  • $\sum_{k=1}^{n} (-1)^{k-1} k^2 = (-1)^{n-1} \frac{n(n+1)}{2}$
  • $\sum_{k=1}^{n} (-1)^{k-1} k^3 = \frac{1}{4} (2n^3+3n^2-1)$ for odd $n$.
• Algebraic Expansion and Simplification: Expanding polynomial expressions and combining like terms.

Step-by-Step Solution

Step 1: Identify the General Term ($T_k$) We need to find a formula for the $k$-th term of the given series: $1^{2}-2 \cdot 3^{2}+3 \cdot 5^{2}-4 \cdot 7^{2}+5 \cdot 9^{2}-\ldots+15 \cdot 29^{2}$. Observing the pattern:

• The coefficient of each term follows the sequence $1, 2, 3, 4, 5, \ldots, 15$. This is simply $k$.
• The base of the squared term follows the sequence of odd numbers: $1, 3, 5, 7, 9, \ldots, 29$. The $k$-th odd number can be expressed as $2k-1$. So, the squared term is $(2k-1)^2$.
• The signs alternate: positive for the 1st, 3rd, 5th, ... terms (odd $k$) and negative for the 2nd, 4th, 6th, ... terms (even $k$). This alternating sign can be represented by $(-1)^{k-1}$.

Therefore, the general term $T_k$ is given by: $T_k = (-1)^{k-1} k (2k-1)^2$ The series ends with the term $15 \cdot 29^2$. Let's check if our general term matches for $k=15$: $T_{15} = (-1)^{15-1} \cdot 15 \cdot (2 \cdot 15 - 1)^2 = (-1)^{14} \cdot 15 \cdot (30-1)^2 = 1 \cdot 15 \cdot 29^2$. This matches. The total number of terms in the series is $N=15$.

Step 2: Expand the General Term into a Polynomial To use the summation formulas for powers of $k$, we need to expand the expression for $T_k$: $k(2k-1)^2 = k(4k^2 - 4k + 1)$ $k(2k-1)^2 = 4k^3 - 4k^2 + k$ So, the general term is: $T_k = (-1)^{k-1} (4k^3 - 4k^2 + k)$

Step 3: Decompose the Sum into Standard Alternating Power Series The sum of the series, $S$, can be written as: $S = \sum_{k=1}^{15} T_k = \sum_{k=1}^{15} (-1)^{k-1} (4k^3 - 4k^2 + k)$ We can split this sum into three separate sums using the linearity of summation: $S = 4 \sum_{k=1}^{15} (-1)^{k-1} k^3 - 4 \sum_{k=1}^{15} (-1)^{k-1} k^2 + \sum_{k=1}^{15} (-1)^{k-1} k$ We need to evaluate each of these sums for $n=15$. Since $n=15$ is an odd number, we will use the specific formulas for odd $n$ where applicable.

Sub-step 3.1: Evaluate $\sum_{k=1}^{15} (-1)^{k-1} k$ Using the formula $\sum_{k=1}^{n} (-1)^{k-1} k = \frac{(-1)^{n-1}(2n+1)+1}{4}$ with $n=15$: $\sum_{k=1}^{15} (-1)^{k-1} k = \frac{(-1)^{15-1}(2 \cdot 15 + 1) + 1}{4} = \frac{(-1)^{14}(30+1) + 1}{4}$ $= \frac{1 \cdot 31 + 1}{4} = \frac{32}{4} = 8$

Sub-step 3.2: Evaluate $\sum_{k=1}^{15} (-1)^{k-1} k^2$ Using the formula $\sum_{k=1}^{n} (-1)^{k-1} k^2 = (-1)^{n-1} \frac{n(n+1)}{2}$ with $n=15$: $\sum_{k=1}^{15} (-1)^{k-1} k^2 = (-1)^{15-1} \frac{15(15+1)}{2} = (-1)^{14} \frac{15 \cdot 16}{2}$ $= 1 \cdot \frac{240}{2} = 120$

Sub-step 3.3: Evaluate $\sum_{k=1}^{15} (-1)^{k-1} k^3$ Using the formula $\sum_{k=1}^{n} (-1)^{k-1} k^3 = \frac{1}{4} (2n^3+3n^2-1)$ for odd $n$, with $n=15$: $\sum_{k=1}^{15} (-1)^{k-1} k^3 = \frac{1}{4} (2 \cdot 15^3 + 3 \cdot 15^2 - 1)$ First, calculate the powers of 15: $15^2 = 225$, $15^3 = 15 \cdot 225 = 3375$. $= \frac{1}{4} (2 \cdot 3375 + 3 \cdot 225 - 1)$ $= \frac{1}{4} (6750 + 675 - 1)$ $= \frac{1}{4} (7425 - 1) = \frac{7424}{4} = 1856$

Step 4: Combine the Results to Find the Total Sum Substitute the calculated values back into the expression for $S$: $S = 4 (1856) - 4 (120) + 8$ $S = 7424 - 480 + 8$ $S = 6944 + 8$ $S = 6952$

Common Mistakes & Tips

• General Term Identification: Ensure the formula for the $k$-th term correctly captures the coefficient, the base of the power, and the alternating sign for all terms.
• Algebraic Errors: Be meticulous when expanding $(2k-1)^2$ and multiplying by $k$. Small errors here propagate through the entire calculation.
• Summation Formula Accuracy: Use the correct formulas for alternating sums, especially noting any differences for odd versus even values of $n$. The formula for $\sum k^3$ is specifically for odd $n$ in the provided list.
• Arithmetic Precision: Calculations with large numbers like $15^3$ require careful arithmetic. Double-checking these calculations is crucial.

Summary The given series is an alternating series with terms involving a polynomial in $k$. By identifying the general term $T_k = (-1)^{k-1} k (2k-1)^2$, we expanded it into a polynomial $T_k = (-1)^{k-1} (4k^3 - 4k^2 + k)$. The sum was then computed by decomposing it into three standard alternating power series: $\sum (-1)^{k-1} k^3$, $\sum (-1)^{k-1} k^2$, and $\sum (-1)^{k-1} k$. Using the appropriate summation formulas for $n=15$ (an odd number), the individual sums were calculated, and then combined to obtain the final sum of the series.

The final answer is $\boxed{6952}$.