Key Concepts and Formulas

• Sum of the first $N$ natural numbers: $\sum_{k=1}^N k = \frac{N(N+1)}{2}$
• Sum of the first $N$ squares: $\sum_{k=1}^N k^2 = \frac{N(N+1)(2N+1)}{6}$
• Sum of the first $N$ cubes: $\sum_{k=1}^N k^3 = \left( \frac{N(N+1)}{2} \right)^2$
• Linearity of Summation: For constants $a, b$ and functions $f(n), g(n)$, $\sum_{n=1}^N (a \cdot f(n) + b \cdot g(n)) = a \sum_{n=1}^N f(n) + b \sum_{n=1}^N g(n)$

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Step-by-Step Solution

Step 1: Simplify the General Term of the Series The general term of the series is given by $T_n = \frac{n(n+1)(2n+1)}{4}$. To evaluate the sum, we first expand the numerator into a polynomial in $n$. This will allow us to use the standard summation formulas for powers of $n$. Expanding the numerator: $n(n+1)(2n+1) = (n^2 + n)(2n+1)$ $= n^2(2n) + n^2(1) + n(2n) + n(1)$ $= 2n^3 + n^2 + 2n^2 + n$ $= 2n^3 + 3n^2 + n$ So, the general term becomes: $T_n = \frac{1}{4}(2n^3 + 3n^2 + n)$

Step 2: Apply Linearity of Summation We need to calculate the sum $\sum\limits_{n = 1}^7 T_n$. Using the linearity of summation, we can distribute the sum and pull out constant factors. $\sum_{n=1}^7 \frac{1}{4}(2n^3 + 3n^2 + n) = \frac{1}{4} \sum_{n=1}^7 (2n^3 + 3n^2 + n)$ $= \frac{1}{4} \left( \sum_{n=1}^7 2n^3 + \sum_{n=1}^7 3n^2 + \sum_{n=1}^7 n \right)$ $= \frac{1}{4} \left( 2\sum_{n=1}^7 n^3 + 3\sum_{n=1}^7 n^2 + \sum_{n=1}^7 n \right)$

Step 3: Evaluate Individual Sums Now, we use the standard summation formulas with $N=7$.

1. Sum of the first 7 natural numbers: $\sum_{n=1}^7 n = \frac{7(7+1)}{2} = \frac{7 \times 8}{2} = \frac{56}{2} = 28$
2. Sum of the first 7 squares: $\sum_{n=1}^7 n^2 = \frac{7(7+1)(2 \times 7 + 1)}{6} = \frac{7 \times 8 \times 15}{6} = \frac{840}{6} = 140$
3. Sum of the first 7 cubes: $\sum_{n=1}^7 n^3 = \left( \frac{7(7+1)}{2} \right)^2 = (28)^2 = 784$

Step 4: Combine Results and Compute the Final Sum Substitute the values of the individual sums back into the expression from Step 2: $\sum_{n=1}^7 T_n = \frac{1}{4} \left( 2\sum_{n=1}^7 n^3 + 3\sum_{n=1}^7 n^2 + \sum_{n=1}^7 n \right)$ $= \frac{1}{4} \left( 2(784) + 3(140) + 28 \right)$ $= \frac{1}{4} \left( 1568 + 420 + 28 \right)$ $= \frac{1}{4} \left( 1988 + 28 \right)$ $= \frac{1}{4} \left( 2016 \right)$ $= 504$

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Common Mistakes & Tips

• Arithmetic Errors: Carefully performing calculations, especially with larger numbers, is crucial. Double-checking additions and multiplications can prevent mistakes.
• Misremembering Formulas: Ensure you have the correct formulas for sums of powers. A common error is mixing up the formulas for sum of squares and sum of cubes.
• Expansion Errors: When expanding polynomials, ensure each term is correctly multiplied. Errors in expansion will propagate through the entire calculation.

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Summary The problem requires evaluating a finite sum whose general term is a cubic polynomial in $n$. The strategy involves expanding the general term into a sum of powers of $n$, applying the linearity property of summation to separate the sum into standard forms, and then using the known formulas for the sum of the first $N$ natural numbers, their squares, and their cubes. Substituting $N=7$ into these formulas and performing the arithmetic yields the final result.

The final answer is $\boxed{504}$.