Key Concepts and Formulas

• Method of Differences: If the first differences of a sequence are in an arithmetic progression, the general term is a quadratic in $n$.
• Summation Formulas:
  • $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
  • $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Step-by-Step Solution

Step 1: Find the general term $a_n$ of the series. We are given the series $5, 8, 14, 23, 35, 50, \ldots$. Let's find the differences between consecutive terms: First differences: $8-5=3$, $14-8=6$, $23-14=9$, $35-23=12$, $50-35=15$. The first differences are $3, 6, 9, 12, 15, \ldots$. This is an arithmetic progression with a common difference of $3$. Since the first differences form an AP, the general term $a_n$ is a quadratic in $n$, i.e., $a_n = An^2 + Bn + C$.

We use the first three terms to find the coefficients $A, B, C$: For $n=1$: $a_1 = A(1)^2 + B(1) + C = A+B+C = 5$ (Equation 1) For $n=2$: $a_2 = A(2)^2 + B(2) + C = 4A+2B+C = 8$ (Equation 2) For $n=3$: $a_3 = A(3)^2 + B(3) + C = 9A+3B+C = 14$ (Equation 3)

Subtracting Equation 1 from Equation 2: $(4A+2B+C) - (A+B+C) = 8 - 5 \Rightarrow 3A+B = 3$ (Equation 4)

Subtracting Equation 2 from Equation 3: $(9A+3B+C) - (4A+2B+C) = 14 - 8 \Rightarrow 5A+B = 6$ (Equation 5)

Subtracting Equation 4 from Equation 5: $(5A+B) - (3A+B) = 6 - 3 \Rightarrow 2A = 3 \Rightarrow A = \frac{3}{2}$.

Substitute $A = \frac{3}{2}$ into Equation 4: $3(\frac{3}{2}) + B = 3 \Rightarrow \frac{9}{2} + B = 3 \Rightarrow B = 3 - \frac{9}{2} = \frac{6-9}{2} = -\frac{3}{2}$.

Substitute $A = \frac{3}{2}$ and $B = -\frac{3}{2}$ into Equation 1: $\frac{3}{2} - \frac{3}{2} + C = 5 \Rightarrow C = 5$.

Thus, the general term is $a_n = \frac{3}{2}n^2 - \frac{3}{2}n + 5 = \frac{3n^2 - 3n + 10}{2}$.

Step 2: Calculate $a_{40}$. Substitute $n=40$ into the general term $a_n$: $a_{40} = \frac{3(40)^2 - 3(40) + 10}{2} = \frac{3(1600) - 120 + 10}{2} = \frac{4800 - 120 + 10}{2} = \frac{4690}{2} = 2345$

Step 3: Calculate $S_{30}$. $S_{30} = \sum_{k=1}^{30} a_k$. We have $a_k = \frac{3k^2 - 3k + 10}{2}$. $S_{30} = \sum_{k=1}^{30} \frac{3k^2 - 3k + 10}{2} = \frac{1}{2} \sum_{k=1}^{30} (3k^2 - 3k + 10)$ $S_{30} = \frac{1}{2} \left( 3\sum_{k=1}^{30} k^2 - 3\sum_{k=1}^{30} k + \sum_{k=1}^{30} 10 \right)$ Using the summation formulas: $\sum_{k=1}^{30} k = \frac{30(30+1)}{2} = \frac{30 \times 31}{2} = 15 \times 31 = 465$. $\sum_{k=1}^{30} k^2 = \frac{30(30+1)(2 \times 30 + 1)}{6} = \frac{30 \times 31 \times 61}{6} = 5 \times 31 \times 61 = 155 \times 61 = 9455$. $\sum_{k=1}^{30} 10 = 10 \times 30 = 300$.

Substitute these values back into the expression for $S_{30}$: $S_{30} = \frac{1}{2} (3 \times 9455 - 3 \times 465 + 300)$ $S_{30} = \frac{1}{2} (28365 - 1395 + 300)$ $S_{30} = \frac{1}{2} (26970 + 300)$ $S_{30} = \frac{1}{2} (27270) = 13635$

Step 4: Calculate $S_{30} - a_{40}$. $S_{30} - a_{40} = 13635 - 2345 = 11290$

Common Mistakes & Tips

• Arithmetic Errors: Double-check all calculations, especially when dealing with large numbers and fractions. A single mistake can lead to an incorrect final answer.
• Formula Application: Ensure the correct summation formulas for $\sum k$ and $\sum k^2$ are used and applied with the correct upper limit $n$.
• General Term Derivation: Be meticulous when solving for the coefficients $A, B, C$ of the quadratic general term. Errors here will propagate.

Summary We first determined the general term $a_n$ of the given series by analyzing the differences between consecutive terms. Since the second differences were constant, we identified $a_n$ as a quadratic in $n$ and solved for its coefficients. Subsequently, we calculated the specific term $a_{40}$ by substituting $n=40$ into the general term. We then computed the sum of the first 30 terms, $S_{30}$, using the standard summation formulas for powers of $n$. Finally, we subtracted $a_{40}$ from $S_{30}$ to obtain the required value.

The final answer is $\boxed{11290}$.