Key Concepts and Formulas

• Method of Differences: If the first differences of a series do not form an AP or GP, but the second differences are constant, the general term ($T_n$) of the series is a quadratic polynomial of the form $T_n = an^2 + bn + c$.
• Summation Formulas:
  • $\sum_{n=1}^{N} 1 = N$
  • $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$
  • $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$
• Linearity of Summation: $\sum_{n=1}^{N} (c_1 f(n) + c_2 g(n)) = c_1 \sum_{n=1}^{N} f(n) + c_2 \sum_{n=1}^{N} g(n)$

Step-by-Step Solution

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

We are given the series $5+11+19+29+41+\ldots$. To find the general term, we will use the method of differences.

First, let's list the terms of the series and calculate the differences between consecutive terms. The terms are: $T_1 = 5, T_2 = 11, T_3 = 19, T_4 = 29, T_5 = 41$.

Calculate the first differences: $D_1 = T_2 - T_1 = 11 - 5 = 6$ $D_2 = T_3 - T_2 = 19 - 11 = 8$ $D_3 = T_4 - T_3 = 29 - 19 = 10$ $D_4 = T_5 - T_4 = 41 - 29 = 12$ The sequence of first differences is $6, 8, 10, 12, \ldots$.

Now, calculate the second differences (differences between consecutive first differences): $D'_1 = D_2 - D_1 = 8 - 6 = 2$ $D'_2 = D_3 - D_2 = 10 - 8 = 2$ $D'_3 = D_4 - D_3 = 12 - 10 = 2$ The sequence of second differences is $2, 2, 2, \ldots$. Since the second differences are constant, the general term ($T_n$) of the series is a quadratic polynomial of the form $T_n = an^2 + bn + c$.

We can find the coefficients $a, b, c$ by setting up a system of equations using the first three terms of the series: For $n=1$: $T_1 = a(1)^2 + b(1) + c = a + b + c = 5 \quad \ldots (1)$ For $n=2$: $T_2 = a(2)^2 + b(2) + c = 4a + 2b + c = 11 \quad \ldots (2)$ For $n=3$: $T_3 = a(3)^2 + b(3) + c = 9a + 3b + c = 19 \quad \ldots (3)$

Subtract equation (1) from equation (2): $(4a + 2b + c) - (a + b + c) = 11 - 5$ $3a + b = 6 \quad \ldots (4)$

Subtract equation (2) from equation (3): $(9a + 3b + c) - (4a + 2b + c) = 19 - 11$ $5a + b = 8 \quad \ldots (5)$

Subtract equation (4) from equation (5): $(5a + b) - (3a + b) = 8 - 6$ $2a = 2$ $a = 1$

Substitute $a=1$ into equation (4): $3(1) + b = 6$ $3 + b = 6$ $b = 3$

Substitute $a=1$ and $b=3$ into equation (1): $1 + 3 + c = 5$ $4 + c = 5$ $c = 1$

Therefore, the general term of the series is $T_n = n^2 + 3n + 1$.

We can verify this formula with the given terms: $T_1 = 1^2 + 3(1) + 1 = 1 + 3 + 1 = 5$ $T_2 = 2^2 + 3(2) + 1 = 4 + 6 + 1 = 11$ $T_3 = 3^2 + 3(3) + 1 = 9 + 9 + 1 = 19$ The formula is correct.

Step 2: Calculate the Sum of the First $20$ Terms ($S_{20}$)

The sum of the first $N$ terms of the series is given by $S_N = \sum_{n=1}^{N} T_n$. Substituting the general term $T_n = n^2 + 3n + 1$, we get: $S_N = \sum_{n=1}^{N} (n^2 + 3n + 1)$ Using the linearity of summation, we can split this into three separate summations: $S_N = \sum_{n=1}^{N} n^2 + 3 \sum_{n=1}^{N} n + \sum_{n=1}^{N} 1$

Now, we apply the standard summation formulas: $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$ $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$ $\sum_{n=1}^{N} 1 = N$

Substitute these formulas into the expression for $S_N$: $S_N = \frac{N(N+1)(2N+1)}{6} + 3 \left( \frac{N(N+1)}{2} \right) + N$

We need to find the sum of the first $20$ terms, so we set $N=20$: $S_{20} = \frac{20(20+1)(2 \cdot 20 + 1)}{6} + 3 \left( \frac{20(20+1)}{2} \right) + 20$ $S_{20} = \frac{20(21)(41)}{6} + 3 \left( \frac{20(21)}{2} \right) + 20$

Now, we calculate each part:

1. The first term: $\frac{20 \cdot 21 \cdot 41}{6} = \frac{420 \cdot 41}{6} = 70 \cdot 41 = 2870$
2. The second term: $3 \cdot \frac{20 \cdot 21}{2} = 3 \cdot (10 \cdot 21) = 3 \cdot 210 = 630$
3. The third term: $20$

Add these values together to find $S_{20}$: $S_{20} = 2870 + 630 + 20$ $S_{20} = 3500 + 20$ $S_{20} = 3520$

Common Mistakes & Tips

• Accuracy in Differences: Errors in calculating the first or second differences will lead to an incorrect general term ($T_n$). Always double-check these calculations.
• Correct Summation Formulas: Ensure you are using the correct standard formulas for the sum of the first $N$ integers and the sum of the first $N$ squares.
• Algebraic Simplification: Be meticulous when substituting $N$ into the sum formula and performing the arithmetic. Factoring common terms before substitution can sometimes reduce errors, but direct calculation is also valid if done carefully.

Summary

The given series is not an arithmetic or geometric progression. By applying the method of differences, we found that the second differences are constant, indicating that the general term $T_n$ is a quadratic polynomial. We determined $T_n = n^2 + 3n + 1$. To find the sum of the first 20 terms, $S_{20}$, we used the linearity of summation and the standard formulas for $\sum n^2$, $\sum n$, and $\sum 1$. Substituting $N=20$ into the derived sum formula, we calculated $S_{20}$ to be 3520.

The final answer is \boxed{3520}, which corresponds to option (D).