### Related Formula textMethod of differences for a sequence: V_n - V_n-1 = T_n$\text{Method of differences for a sequence: } V_n - V_{n-1} = T_n$ ### Core Logic The base terms inside the squares form a sequence: 1, 4, 8, 13, 19, 26 dots$1, 4, 8, 13, 19, 26 \dots$ The differences between consecutive terms are: 3, 4, 5, 6, 7 dots$3, 4, 5, 6, 7 \dots$ Since the first differences are an Arithmetic Progression, the general term of the inner sequence is a quadratic in n$n$: T_n = an^2 + bn + c$T_n = an^2 + bn + c$. Using n=1$n=1$: 1 = a + b + c$1 = a + b + c$ Using n=2$n=2$: 4 = 4a + 2b + c$4 = 4a + 2b + c$ Using n=3$n=3$: 8 = 9a + 3b + c$8 = 9a + 3b + c$ Solving this system: (4a + 2b + c) - (a + b + c) = 3 Rightarrow 3a + b = 3$(4a + 2b + c) - (a + b + c) = 3 \Rightarrow 3a + b = 3$ (9a + 3b + c) - (4a + 2b + c) = 4 Rightarrow 5a + b = 4$(9a + 3b + c) - (4a + 2b + c) = 4 \Rightarrow 5a + b = 4$ Subtracting these gives: 2a = 1 Rightarrow a = 1/2$2a = 1 \Rightarrow a = 1/2$. Then 3(1/2) + b = 3 Rightarrow b = 3/2$3(1/2) + b = 3 \Rightarrow b = 3/2$. Finally 1/2 + 3/2 + c = 1 Rightarrow c = -1$1/2 + 3/2 + c = 1 \Rightarrow c = -1$. Inner sequence T_n = frac12n^2 + frac32n - 1$T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1$. ### Step 1: Calculating alpha structure The series is alpha = sum_n=1^10 (T_n)^2$\alpha = \sum_{n=1}^{10} (T_n)^2$. 4alpha = sum_n=1^10 4left(fracn^2 + 3n - 22right)^2 = sum_n=1^10 (n^2 + 3n - 2)^2$4\alpha = \sum_{n=1}^{10} 4\left(\frac{n^2 + 3n - 2}{2}\right)^2 = \sum_{n=1}^{10} (n^2 + 3n - 2)^2$ Expand the squared trinomial: (n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4$(n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4$ ### Step 2: Applying given target relation We are given beta = sum_n=1^10 n^4$\beta = \sum_{n=1}^{10} n^4$. So, 4alpha - beta = sum_n=1^10 (n^4 + 6n^3 + 5n^2 - 12n + 4) - sum_n=1^10 n^4$4\alpha - \beta = \sum_{n=1}^{10} (n^4 + 6n^3 + 5n^2 - 12n + 4) - \sum_{n=1}^{10} n^4$ 4alpha - beta = sum_n=1^10 (6n^3 + 5n^2 - 12n + 4)$4\alpha - \beta = \sum_{n=1}^{10} (6n^3 + 5n^2 - 12n + 4)$ ### Step 3: Calculating summation limits Evaluate each standard summation up to n=10$n=10$: sum n^3 = (10 times 11 / 2)^2 = 55^2 = 3025$\sum n^3 = (10 \times 11 / 2)^2 = 55^2 = 3025$ sum n^2 = (10 times 11 times 21) / 6 = 385$\sum n^2 = (10 \times 11 \times 21) / 6 = 385$ sum n = (10 times 11) / 2 = 55$\sum n = (10 \times 11) / 2 = 55$ sum 4 = 40$\sum 4 = 40$ 4alpha - beta = 6(3025) + 5(385) - 12(55) + 40$4\alpha - \beta = 6(3025) + 5(385) - 12(55) + 40$ = 18150 + 1925 - 660 + 40 = 19455$= 18150 + 1925 - 660 + 40 = 19455$ We are given 4alpha - beta = 55k + 40$4\alpha - \beta = 55k + 40$. 19455 = 55k + 40$19455 = 55k + 40$ 19415 = 55k$19415 = 55k$ k = frac1941555 = 353$k = \frac{19415}{55} = 353$ ### Pattern Recognition Recognizing arithmetic progressions in the first-order differences immediately specifies a quadratic general term An^2+Bn+C$An^2+Bn+C$. Expanding and cancelling highest-order summation terms drastically simplifies standard sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$ textIf p, q, r text are in G.P. then q^2 = pr$\text{If } p, q, r \text{ are in G.P. then } q^2 = pr$ ### Core Logic Let the A.P. be a, a+d, a+2d, dots$a, a+d, a+2d, \dots$ Given a=1$a=1$, the 2^textnd, 8^textth$2^{\text{nd}}, 8^{\text{th}}$, and 44^textth$44^{\text{th}}$ terms are: T_2 = 1 + d$T_2 = 1 + d$ T_8 = 1 + 7d$T_8 = 1 + 7d$ T_44 = 1 + 43d$T_{44} = 1 + 43d$ These terms are in G.P., so: (1+7d)^2 = (1+d)(1+43d)$(1+7d)^2 = (1+d)(1+43d)$ 1 + 14d + 49d^2 = 1 + 44d + 43d^2$1 + 14d + 49d^2 = 1 + 44d + 43d^2$ 6d^2 - 30d = 0 implies 6d(d - 5) = 0$6d^2 - 30d = 0 \implies 6d(d - 5) = 0$ Since it is a non-constant A.P., d neq 0$d \neq 0$, so d = 5$d = 5$. Sum of first 20 terms: S_20 = frac202[2(1) + (20-1)5]$S_{20} = \frac{20}{2}[2(1) + (20-1)5]$ S_20 = 10[2 + 19(5)] = 10[2 + 95] = 970$S_{20} = 10[2 + 19(5)] = 10[2 + 95] = 970$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Core Logic General term T_r = fracrr^4 - 3r^2 + 1$T_r = \frac{r}{r^4 - 3r^2 + 1}$. Factorize the denominator: r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2$r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2$ = (r^2 - r - 1)(r^2 + r - 1)$= (r^2 - r - 1)(r^2 + r - 1)$ ### Step 1: Partial Fractions T_r = fracr(r^2 - r - 1)(r^2 + r - 1)$T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}$ Notice that (r^2 + r - 1) - (r^2 - r - 1) = 2r$(r^2 + r - 1) - (r^2 - r - 1) = 2r$. T_r = frac12 left[ frac2r(r^2 - r - 1)(r^2 + r - 1) right] = frac12 left[ frac1r^2 - r - 1 - frac1r^2 + r - 1 right]$T_r = \frac{1}{2} \left[ \frac{2r}{(r^2 - r - 1)(r^2 + r - 1)} \right] = \frac{1}{2} \left[ \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right]$ ### Step 2: Telescoping Sum Sum S = sum_r=1^10 T_r$S = \sum_{r=1}^{10} T_r$. The terms will telescope because the second term for r$r$ is identical to the first term for r+1$r+1$. (Let v_r = r^2 - r - 1$v_r = r^2 - r - 1$, then v_r+1 = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1$v_{r+1} = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1$). S = frac12 left[ frac11^2 - 1 - 1 - frac110^2 + 10 - 1 right]$S = \frac{1}{2} \left[ \frac{1}{1^2 - 1 - 1} - \frac{1}{10^2 + 10 - 1} \right]$ S = frac12 left[ frac1-1 - frac1109 right] = frac12 left[ -1 - frac1109 right] = -frac55109$S = \frac{1}{2} \left[ \frac{1}{-1} - \frac{1}{109} \right] = \frac{1}{2} \left[ -1 - \frac{1}{109} \right] = -\frac{55}{109}$ ### Pattern Recognition Expressions like r^4 + kr^2 + 1$r^4 + kr^2 + 1$ can be factorized by completing the square to create a difference of two squares. This setup invariably leads to a telescoping series. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series