Key Concepts and Formulas

• Method of Differences: Applicable to series where the differences between consecutive terms, or the differences of these differences, form an AP. If the second differences are constant, the $n$-th term ($T_n$) is a quadratic in $n$.
• Summation of Powers:
  • $\sum_{k=1}^n k = \frac{n(n+1)}{2}$
  • $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
• Sum of a Series: If $T_n = An^2 + Bn + C$, then $S_n = \sum_{k=1}^n T_k$ will be a cubic polynomial in $n$.

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Step-by-Step Solution

Step 1: Determine the General Term ($T_n$) of the Series

We are given the series $S_n = 4 + 11 + 21 + 34 + 50 + \ldots$. Let the $n$-th term be $T_n$. The first few terms are: $T_1 = 4$ $T_2 = 11$ $T_3 = 21$ $T_4 = 34$ $T_5 = 50$

We find the first differences between consecutive terms: $T_2 - T_1 = 11 - 4 = 7$ $T_3 - T_2 = 21 - 11 = 10$ $T_4 - T_3 = 34 - 21 = 13$ $T_5 - T_4 = 50 - 34 = 16$ The sequence of first differences is $7, 10, 13, 16, \ldots$.

Now, we find the second differences: $10 - 7 = 3$ $13 - 10 = 3$ $16 - 13 = 3$ Since the second differences are constant (equal to 3), the first differences form an Arithmetic Progression (AP), and the general term $T_n$ is a quadratic polynomial of the form $T_n = An^2 + Bn + C$.

We use the first three terms to set up equations for $A, B, C$: For $n=1$: $A(1)^2 + B(1) + C = A+B+C = 4 \quad \ldots (1)$ For $n=2$: $A(2)^2 + B(2) + C = 4A+2B+C = 11 \quad \ldots (2)$ For $n=3$: $A(3)^2 + B(3) + C = 9A+3B+C = 21 \quad \ldots (3)$

Subtracting (1) from (2): $(4A+2B+C) - (A+B+C) = 11 - 4 \implies 3A+B = 7 \quad \ldots (4)$

Subtracting (2) from (3): $(9A+3B+C) - (4A+2B+C) = 21 - 11 \implies 5A+B = 10 \quad \ldots (5)$

Subtracting (4) from (5): $(5A+B) - (3A+B) = 10 - 7 \implies 2A = 3 \implies A = \frac{3}{2}$

Substitute $A = \frac{3}{2}$ into (4): $3\left(\frac{3}{2}\right) + B = 7 \implies \frac{9}{2} + B = 7 \implies B = 7 - \frac{9}{2} = \frac{14-9}{2} = \frac{5}{2}$

Substitute $A = \frac{3}{2}$ and $B = \frac{5}{2}$ into (1): $\frac{3}{2} + \frac{5}{2} + C = 4 \implies \frac{8}{2} + C = 4 \implies 4 + C = 4 \implies C = 0$

So, the general term is $T_n = \frac{3}{2}n^2 + \frac{5}{2}n = \frac{n(3n+5)}{2}$.

Step 2: Find the Sum of the Series ($S_n$)

The sum $S_n$ is the sum of the first $n$ terms: $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n \left(\frac{3}{2}k^2 + \frac{5}{2}k\right)$ We can split the summation and factor out constants: $S_n = \frac{3}{2}\sum_{k=1}^n k^2 + \frac{5}{2}\sum_{k=1}^n k$

Using the standard summation formulas: $S_n = \frac{3}{2}\left(\frac{n(n+1)(2n+1)}{6}\right) + \frac{5}{2}\left(\frac{n(n+1)}{2}\right)$ $S_n = \frac{n(n+1)(2n+1)}{4} + \frac{5n(n+1)}{4}$ Factor out the common term $\frac{n(n+1)}{4}$: $S_n = \frac{n(n+1)}{4} \left((2n+1) + 5\right)$ $S_n = \frac{n(n+1)}{4} (2n+6)$ $S_n = \frac{n(n+1) \cdot 2(n+3)}{4}$ $S_n = \frac{n(n+1)(n+3)}{2}$

Step 3: Calculate $S_{29} - S_9$

We need to calculate $S_{29}$ and $S_9$ using the formula $S_n = \frac{n(n+1)(n+3)}{2}$.

For $n=29$: $S_{29} = \frac{29(29+1)(29+3)}{2} = \frac{29 \times 30 \times 32}{2}$ $S_{29} = 29 \times 15 \times 32 = 29 \times 480$ $S_{29} = 13920$

For $n=9$: $S_9 = \frac{9(9+1)(9+3)}{2} = \frac{9 \times 10 \times 12}{2}$ $S_9 = 9 \times 5 \times 12 = 9 \times 60$ $S_9 = 540$

Now, we find the difference: $S_{29} - S_9 = 13920 - 540 = 13380$

Step 4: Calculate the Final Expression

We are asked to find $\frac{1}{60}(S_{29} - S_9)$. $\frac{1}{60}(S_{29} - S_9) = \frac{1}{60}(13380)$ $\frac{13380}{60} = \frac{1338}{6}$

Performing the division: $1338 \div 6 = 223$

The value of $\frac{1}{60}(S_{29} - S_9)$ is 223.

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Common Mistakes & Tips

• Arithmetic Errors: Be extremely careful with calculations, especially multiplications and subtractions involving larger numbers.
• Incorrect General Term: Ensure that the method of differences is applied correctly and that the derived $T_n$ matches the initial terms of the series.
• Formula Application: Double-check the standard summation formulas for $k$ and $k^2$.

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Summary

This problem required identifying the pattern in the series using the method of differences to find the general term $T_n$. Since the second differences were constant, $T_n$ was a quadratic in $n$. We then used the standard formulas for the sum of powers of integers to derive the sum of the series $S_n$. Finally, we substituted the required values of $n$ into the formula for $S_n$, calculated the difference $S_{29} - S_9$, and then divided by 60 to obtain the final answer.

The final answer is $\boxed{223}$.