Question:

Step 1: Identify the General Term The given series is $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \ldots$. Let $T_n$ denote the $n$-th term of the series. The terms are: $T_1 = 1$ $T_2 = 2 \times 3$ $T_3 = 3 \times 5$ $T_4 = 4 \times 7$ Observe the pattern for the $n$-th term. The first factor is $n$. The second factor is an odd number. For $T_n$, the second factor is $2n-1$. Let's verify this pattern: For $n=1$: $T_1 = 1 \times (2 \cdot 1 - 1) = 1 \times 1 = 1$. This matches the first term. For $n=2$: $T_2 = 2 \times (2 \cdot 2 - 1) = 2 \times 3 = 6$. This matches the second term. For $n=3$: $T_3 = 3 \times (2 \cdot 3 - 1) = 3 \times 5 = 15$. This matches the third term. For $n=4$: $T_4 = 4 \times (2 \cdot 4 - 1) = 4 \times 7 = 28$. This matches the fourth term. Thus, the general term for the series is $T_n = n(2n-1)$. Step 2: Expand the General Term The general term $T_n$ can be expanded as: $T_n = n(2n-1) = 2n^2 - n$. Step 3: Apply Summation Formula The sum of the series up to the $11^{\text{th}}$ term, denoted as $S_{11}$, is given by: $$S_{11} = \sum_{n=1}^{11} T_n = \sum_{n=1}^{11} (2n^2 - n)$$ This can be split into two separate summations: $$S_{11} = 2 \sum_{n=1}^{11} n^2 - \sum_{n=1}^{11} n$$ Step 4: Use Standard Summation Formulas The standard formulas for the sum of the first $k$ natural numbers and the sum of the squares of the first $k$ natural numbers are: $$\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$$ $$\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$$ For $k=11$: $$\sum_{n=1}^{11} n = \frac{11(11+1)}{2} = \frac{11 \times 12}{2} = 11 \times 6 = 66$$ $$\sum_{n=1}^{11} n^2 = \frac{11(11+1)(2 \times 11+1)}{6} = \frac{11 \times 12 \times 23}{6} = 11 \times 2 \times 23 = 506$$ Step 5: Calculate the Final Sum Substitute the calculated summation values back into the expression for $S_{11}$: $$S_{11} = 2(506) - 66$$ $$S_{11} = 1012 - 66$$ $$S_{11} = 946$$