If the sum of the first 10 terms of the series frac4 cdot 11 + 4 cdot 1^4 + frac4 cdot 21 + 4 cdot 2^4 + frac4 cdot 31 + 4 cdot 3^4 + dots$\frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \dots$ is fracmathrmmmathrmn$\frac{\mathrm{m}}{\mathrm{n}}$, where gcd(mathrmm, mathrmn) = 1$\gcd(\mathrm{m}, \mathrm{n}) = 1$, then mathrmm + mathrmn$\mathrm{m} + \mathrm{n}$ is equal to ____________.

### Related Formula Characteristic equation method for standard second-order linear homogeneous recurrence updates: 2x^2 - 5x + 3 = 0$$2x^2 - 5x + 3 = 0$$ ### Core Logic Solving the characteristic equation gives roots x = 1$x = 1$ and x = frac32$x = \frac{3}{2}$. The general solution takes the form: a_n = A(1)^n + Bleft(frac32right)^n$$a_n = A(1)^n + B\left(\frac{3}{2}\right)^n$$ ### Step 1: Evaluating Sequence Parameters Using boundary conditions: For n = 0 implies A + B = 0$n = 0 \implies A + B = 0$ For n = 1 implies A + frac32B = frac12$n = 1 \implies A + \frac{3}{2}B = \frac{1}{2}$ Solving this simple linear system gives B = 1$B = 1$ and A = -1$A = -1$. Thus, the explicit sequence formula is: a_n = -1 + left(frac32right)^n$$a_n = -1 + \left(\frac{3}{2}\right)^n$$ ### Step 2: Summing the Target Range sum_k = 1^100 a_k = sum_k = 1^100 (-1) + sum_k = 1^100 left(frac32right)^k$$\sum_{k = 1}^{100} a_k = \sum_{k = 1}^{100} (-1) + \sum_{k = 1}^{100} \left(\frac{3}{2}\right)^k$$ = -100 + fracfrac32left[left(frac32right)^100 - 1right]frac32 - 1 = -100 + 3left[left(frac32right)^100 - 1right]$$= -100 + \frac{\frac{3}{2}\left[\left(\frac{3}{2}\right)^{100} - 1\right]}{\frac{3}{2} - 1} = -100 + 3\left[\left(\frac{3}{2}\right)^{100} - 1\right]$$ = 3a_100 - 100$$= 3a_{100} - 100$$ ### Pattern Recognition Characteristic roots directly decouple second-order linear loop progressions into basic combinations of clean geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula Standard Arithmetic Progression summation template: S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Given structural constraints: T_25 = a + 24d = frac120$$T_{25} = a + 24d = \frac{1}{20}$$ 20 cdot frac252left[a + frac120right] = 13 implies a = frac1500$$20 \cdot \frac{25}{2}\left[a + \frac{1}{20}\right] = 13 \implies a = \frac{1}{500}$$ ### Step 1: Finding Parameters and Indices Substituting a = frac1500$a = \frac{1}{500}$ back into a + 24d = frac120$a + 24d = \frac{1}{20}$ gives d = frac1500$d = \frac{1}{500}$. Using the formula for T_m$T_m$: T_m = a + (m-1)d = frac1500 + fracm-1500 = frac125 implies m = 20$$T_m = a + (m-1)d = \frac{1}{500} + \frac{m-1}{500} = \frac{1}{25} \implies m = 20$$ ### Step 2: Computing the Target Segment Sum For m = 20$m = 20$, the target expression becomes: 5(20) sum_r=20^40 T_r = 100 cdot frac212 [T_20 + T_40]$$5(20) \sum_{r=20}^{40} T_r = 100 \cdot \frac{21}{2} [T_{20} + T_{40}]$$ Evaluating the values gives exactly 126. ### Pattern Recognition When a = d$a = d$, the expressions simplify directly to basic multiples of the index position (T_n = n cdot d$T_n = n \cdot d$), cutting down calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula For series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic form: T_n = an^2 + bn + c$$T_n = an^2 + bn + c$$ ### Core Logic Analyze successive first-order differences of the terms [cite: 1293, 1294]: textSeries: 1, quad 3, quad 11, quad 25, quad 45, quad 71$$\text{Series: } 1, \quad 3, \quad 11, \quad 25, \quad 45, \quad 71$$ [cite: 1293] textDifferences: 2, quad 8, quad 14, quad 20, quad 26$$\text{Differences: } 2, \quad 8, \quad 14, \quad 20, \quad 26$$ [cite: 1294] Since the differences grow uniformly by 6, they reside in an A.P. [cite: 1294] Thus, set up the general term system [cite: 1296, 1298]: - T_1 = a + b + c = 1$T_1 = a + b + c = 1$ - T_2 = 4a + 2b + c = 3$T_2 = 4a + 2b + c = 3$ - T_3 = 9a + 3b + c = 11$T_3 = 9a + 3b + c = 11$ Solving the linear equations simultaneously yields [cite: 1299]: a = 3, quad b = -7, quad c = 5$$a = 3, \quad b = -7, \quad c = 5$$ [cite: 1299] ### Step 1: Summing the Series The general term is [cite: 1300]: T_n = 3n^2 - 7n + 5$$T_n = 3n^2 - 7n + 5$$ [cite: 1300] Evaluate the summation for n=20$n=20$ terms [cite: 1302]: S_20 = sum_n=1^20 (3n^2 - 7n + 5) = 3sum_n=1^20 n^2 - 7sum_n=1^20 n + sum_n=1^20 5$$S_{20} = \sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\sum_{n=1}^{20} n^2 - 7\sum_{n=1}^{20} n + \sum_{n=1}^{20} 5$$ [cite: 1302] Substitute standard sequence formulas [cite: 1302]: S_20 = 3 cdot left(frac20 cdot 21 cdot 416right) - 7 cdot left(frac20 cdot 212right) + 5(20)$$S_{20} = 3 \cdot \left(\frac{20 \cdot 21 \cdot 41}{6}\right) - 7 \cdot \left(\frac{20 \cdot 21}{2}\right) + 5(20)$$ [cite: 1302] = 8610 - 1470 + 100 = 7240$$= 8610 - 1470 + 100 = 7240$$ [cite: 1302] ### Pattern Recognition When first-order differences form a regular arithmetic line, the original function is exactly quadratic. Identify coefficients using small terms quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula For a geometric sequence configuration with first term a$a$ and common ratio r$r$: a_n = a cdot r^n-1$$a_n = a \cdot r^{n-1}$$ ### Core Logic Convert information markers using parameter notations [cite: 1372, 1373, 1376]: a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 implies ar^3 = 27$$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 \implies ar^3 = 27$$ [cite: 1373, 1374] From second expression block data [cite: 1376]: a_2 + a_4 = ar + ar^3 = frac1114$$a_2 + a_4 = ar + ar^3 = \frac{111}{4}$$ [cite: 1376] Substitute ar^3 = 27$ar^3 = 27$ directly into the linear equation block [cite: 1376]: ar + 27 = frac1114 implies ar = frac1114 - 27 = frac34$$ar + 27 = \frac{111}{4} \implies ar = \frac{111}{4} - 27 = \frac{3}{4}$$ [cite: 1376] ### Step 1: Finding parameters a and r Divide the calculated components to evaluate the ratio [cite: 1387]: fracar^3ar = frac273/4 implies r^2 = 36 implies r = 6$$\frac{ar^3}{ar} = \frac{27}{3/4} \implies r^2 = 36 \implies r = 6$$ [cite: 1387] (Choose +6$+6$ because terms must stay strictly positive [cite: 655]). Find value for first base variable a$a$ [cite: 1389]: a(6) = frac34 implies a = frac18$$a(6) = \frac{3}{4} \implies a = \frac{1}{8}$$ [cite: 1389] ### Step 2: Sum configuration resolving Now compute targeted expansion expression value [cite: 1390]: 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$$24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$$ [cite: 1390] = 24 cdot left(frac18right) cdot (1 + 6 + 36) = 3 cdot 43 = 129$$= 24 \cdot \left(\frac{1}{8}\right) \cdot (1 + 6 + 36) = 3 \cdot 43 = 129$$ [cite: 1390, 1391] ### Pattern Recognition Product entries like a_3 a_5 = a_4^2$a_3 a_5 = a_4^2$ help identify the central term index value quickly in symmetric geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Core Logic The general term T_r$T_r$ of the series can be written as: T_r = frac4rr^4 + 3r^2 + 4$$T_r = \frac{4r}{r^4 + 3r^2 + 4}$$ Let's factorize the denominator by completing the square metric: r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$$r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$$ Using the difference of squares identity A^2 - B^2 = (A-B)(A+B)$A^2 - B^2 = (A-B)(A+B)$: r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$$r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$$ ### Step 1: Partial Fraction Decomposition Express T_r$T_r$ using partial fractions split: T_r = frac4r(r^2 - r + 2)(r^2 + r + 2) = 2 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right]$$T_r = \frac{4r}{(r^2 - r + 2)(r^2 + r + 2)} = 2 \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$$ Notice that if we define V(r) = r^2 - r + 2$V(r) = r^2 - r + 2$, then V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$. Thus, T_r = 2big[V(r) - V(r+1)big]$T_r = 2\big[V(r) - V(r+1)\big]$, which sets up a clear telescoping sum formulation. ### Step 2: Evaluating the Sum of 20 Terms Summing from r = 1$r = 1$ to 20$20$: S_20 = sum_r=1^20 T_r = 2 sum_r=1^20 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right]$$S_{20} = \sum_{r=1}^{20} T_r = 2 \sum_{r=1}^{20} \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$$ = 2 left[ left(frac12 - frac14right) + left(frac14 - frac18right) + dots + left(frac120^2 - 20 + 2 - frac120^2 + 20 + 2right) right]$$= 2 \left[ \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \dots + \left(\frac{1}{20^2 - 20 + 2} - \frac{1}{20^2 + 20 + 2}\right) \right]$$ All sequential middle terms cancel completely, leaving only first and final values: S_20 = 2 left[ frac12 - frac1422 right] = 1 - frac1211 = frac210211$$S_{20} = 2 \left[ \frac{1}{2} - \frac{1}{422} \right] = 1 - \frac{1}{211} = \frac{210}{211}$$ Since 210$210$ and 211$211$ are coprime, m = 210$m = 210$ and n = 211$n = 211$. ### Step 3: Calculating m + n Combining both values: m + n = 210 + 211 = 421$$m + n = 210 + 211 = 421$$ ### Pattern Recognition The polynomial factorization r^4 + a^2r^2 + b^4$r^4 + a^2r^2 + b^4$ is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series