### Related Formula Summation Identities: sum r = fracn(n+1)2, quad sum r^2 = fracn(n+1)(2n+1)6$\sum r = \frac{n(n+1)}{2}, \quad \sum r^2 = \frac{n(n+1)(2n+1)}{6}$ ### Core Logic Split the 40-term series into two sub-series of 20 terms each: Series 1 (squared terms at positions 1, 3, 5... wait, positions are odd numbers whose base squares are odd): 1^2 + 5^2 + 9^2 + dots$1^2 + 5^2 + 9^2 + \dots$ upto 20 terms. General term T_r = (4r - 3)^2$T_r = (4r - 3)^2$. Series 2 (linear terms at positions 2, 4, 6...): 3 + 7 + 11 + dots$3 + 7 + 11 + \dots$ upto 20 terms. General term t_r = (4r - 1)$t_r = (4r - 1)$. ### Step 1: Formulate Total Sigma Expression textSum = sum_r=1^20 left[ (4r - 3)^2 + (4r - 1) right]$\text{Sum} = \sum_{r=1}^{20} \left[ (4r - 3)^2 + (4r - 1) \right]$ textSum = sum_r=1^20 (16r^2 - 24r + 9 + 4r - 1) = sum_r=1^20 (16r^2 - 20r + 8)$\text{Sum} = \sum_{r=1}^{20} (16r^2 - 24r + 9 + 4r - 1) = \sum_{r=1}^{20} (16r^2 - 20r + 8)$ textSum = 16sum_r=1^20 r^2 - 20sum_r=1^20 r + 8sum_r=1^20 1$\text{Sum} = 16\sum_{r=1}^{20} r^2 - 20\sum_{r=1}^{20} r + 8\sum_{r=1}^{20} 1$ ### Step 2: Arithmetic Evaluation sum_r=1^20 r^2 = frac20 times 21 times 416 = 2870$\sum_{r=1}^{20} r^2 = \frac{20 \times 21 \times 41}{6} = 2870$ sum_r=1^20 r = frac20 times 212 = 210$\sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210$ textSum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$\text{Sum} = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$ ### Pattern Recognition When dealing with interlaced series, pairing terms adjacent to each other simplifies the degree of general expressions into manageable standard summation polynomials. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequence and Series