Key Concepts and Formulas

• Series Summation: Understanding how to express the sum of $n$ terms ($S_n$) in relation to the sum of $(n-1)$ terms ($S_{n-1}$) and the $n$-th term ($T_n$): $S_n = S_{n-1} + T_n$.
• Pattern Recognition: Identifying the rule for terms in a given series based on their position (index).
• Conditional Formulas: Applying different formulas or logic based on whether a variable (like $n$) is even or odd.

Step-by-Step Solution

Step 1: Analyze the given series and the formula for even $n$. The series is given by ${1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + \dots$ We are given that the sum of the first $n$ terms, $S_n$, is ${{n{{(n + 1)}^2}} \over 2}$ when $n$ is an even number.

Step 2: Determine the general form of the $k$-th term ($T_k$). Let's examine the terms:

• The 1st term ($k=1$, odd): $1^2$
• The 2nd term ($k=2$, even): $2 \cdot 2^2$
• The 3rd term ($k=3$, odd): $3^2$
• The 4th term ($k=4$, even): $2 \cdot 4^2$
• The 5th term ($k=5$, odd): $5^2$

From this pattern, we can deduce that the $k$-th term, $T_k$, is:

• $T_k = k^2$ if $k$ is odd.
• $T_k = 2 \cdot k^2$ if $k$ is even.

Step 3: Set up the calculation for $S_n$ when $n$ is odd. We need to find the sum of the first $n$ terms when $n$ is an odd number. We can express $S_n$ as the sum of the first $(n-1)$ terms plus the $n$-th term: $S_n = S_{n-1} + T_n$ Since $n$ is odd, $(n-1)$ is an even number. This is crucial because we have a formula for $S_k$ when $k$ is even.

Step 4: Calculate $S_{n-1}$ using the given formula for even terms. Since $(n-1)$ is even, we can use the provided formula $S_k = \frac{k(k+1)^2}{2}$ by substituting $k = (n-1)$: $S_{n-1} = \frac{(n-1)((n-1)+1)^2}{2}$ $S_{n-1} = \frac{(n-1)(n)^2}{2}$ $S_{n-1} = \frac{n^2(n-1)}{2}$

Step 5: Determine the $n$-th term ($T_n$) for odd $n$. As established in Step 2, when the index $n$ is odd, the $n$-th term of the series is $T_n = n^2$.

Step 6: Substitute the expressions for $S_{n-1}$ and $T_n$ into the equation for $S_n$. Now, we combine the results from Step 4 and Step 5: $S_n = S_{n-1} + T_n$ $S_n = \frac{n^2(n-1)}{2} + n^2$

Step 7: Simplify the expression for $S_n$. To simplify, we find a common denominator and combine the terms: $S_n = \frac{n^2(n-1)}{2} + \frac{2n^2}{2}$ $S_n = \frac{n^2(n-1) + 2n^2}{2}$ Factor out $n^2$ from the numerator: $S_n = \frac{n^2((n-1) + 2)}{2}$ Simplify the expression inside the parentheses: $S_n = \frac{n^2(n+1)}{2}$

This is the formula for the sum of the first $n$ terms when $n$ is odd.

Common Mistakes & Tips

• Confusing indices: Be careful to distinguish between the general term index $k$ and the specific number of terms $n$.
• Algebraic errors in simplification: When combining fractions or factoring, ensure all steps are accurate. For example, correctly expanding or simplifying $(n-1)+2$.
• Applying the correct formula: Remember that the given formula for $S_n$ is only valid for even $n$. For odd $n$, we use the relationship $S_n = S_{n-1} + T_n$, where $S_{n-1}$ uses the even formula and $T_n$ follows the series' term rule for odd indices.

Summary

To find the sum of the first $n$ terms when $n$ is odd, we utilized the relationship $S_n = S_{n-1} + T_n$. Since $n$ is odd, $n-1$ is even, allowing us to use the provided formula for $S_{n-1}$. We also identified that for an odd index $n$, the $n$-th term is $T_n = n^2$. By substituting these into the sum relation and simplifying, we arrived at the formula for $S_n$ when $n$ is odd.

The derived sum is ${{{n^2}(n + 1)} \over 2}$.

Comparing this with the options: (A) ${\left[ {{{n(n + 1)} \over 2}} \right]^2} = \frac{n^2(n+1)^2}{4}$ (B) ${{{n^2}(n + 1)} \over 2}$ (C) ${{n{{(n + 1)}^2}} \over 4}$ (D) $\,{{3n(n + 1)} \over 2}$

Our result matches option (B).

The final answer is \boxed{(B)}.