### Related Formula textTotal Sum = alpha + beta$\text{Total Sum} = \alpha + \beta$ ### Core Logic Factor out common fractions from the even terms component (beta$\beta$) to represent it as a scalar multiple of the universal sum sequence. ### Step 1: Simplify the Even Terms Series beta = frac12^4 + frac14^4 + frac16^4 + dots = frac12^4 left( frac11^4 + frac12^4 + frac13^4 + dots right)$\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots = \frac{1}{2^4} \left( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots \right)$ beta = frac116 left( fracpi^490 right)$\beta = \frac{1}{16} \left( \frac{\pi^4}{90} \right)$ ### Step 2: Express Alpha by Remainder Deduction Since total sum equals alpha + beta$\alpha + \beta$: alpha = textTotal Sum - beta = fracpi^490 - frac116 left( fracpi^490 right) = frac1516 left( fracpi^490 right)$\alpha = \text{Total Sum} - \beta = \frac{\pi^4}{90} - \frac{1}{16} \left( \frac{\pi^4}{90} \right) = \frac{15}{16} \left( \frac{\pi^4}{90} \right)$ ### Step 3: Compute the Relative Ratio 決fracalphabeta = fracfrac1516 left( fracpi^490 right)frac116 left( fracpi^490 right) = 15$決\frac{\alpha}{\beta} = \frac{\frac{15}{16} \left( \frac{\pi^4}{90} \right)}{\frac{1}{16} \left( \frac{\pi^4}{90} \right)} = 15$ ### Pattern Recognition For alternating p-series powers like $ ### Pattern Recognition For alternating p-series powers like $\sum n^{-p}, the even component fractions always condense via factor steps to $, the even component fractions always condense via factor steps to $2^{-p} \cdot S_{\text{total}}$, decoupling power values cleanly from final simple scalar quotients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of an infinite geometric progression with |r| < 1$|r| < 1$: S_infty = fraca1 - r$S_{\infty} = \frac{a}{1 - r}$ ### Core Logic From the definition of the relation: log_e y = x log_eleft(frac25 ight) implies log_e y = log_eleft(frac25right)^x implies y = left(frac25 ight)^x$\log_e y = x \log_e\left(\frac{2}{5} ight) \implies \log_e y = \log_e\left(\frac{2}{5}\right)^x \implies y = \left(\frac{2}{5} ight)^x$

Since x in S = \0, 1, 2, 3, dots\$x \in S = \{0, 1, 2, 3, \dots\}$, the output values of y$y$ represent elements of the range. ### Step 1: Compute Infinite Sum Generating elements by plugging in values of x$x$: For x = 0 implies y = 1$x = 0 \implies y = 1$ For x = 1 implies y = frac25$x = 1 \implies y = \frac{2}{5}$ For x = 2 implies y = left(frac25 ight)^2$x = 2 \implies y = \left(\frac{2}{5} ight)^2$ Sum of elements in the range: textSum = 1 + left(frac25 ight)^1 + left(frac25 ight)^2 + dots = frac11 - frac25 = frac53$\text{Sum} = 1 + \left(\frac{2}{5} ight)^1 + \left(\frac{2}{5} ight)^2 + \dots = \frac{1}{1 - \frac{2}{5}} = \frac{5}{3}$ ### Pattern Recognition Convert log equations into standard exponential equations right away. A variable index belonging to whole numbers indicates an infinite GP summation scenario. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 11 Mathematics: Relations and Functions

### Related Formula Symmetry identity rule inside Arithmetic Progressions: a_k + a_n-k+1 = a_1 + a_n$a_k + a_{n-k+1} = a_1 + a_n$ ### Core Logic Group matching paired steps equidistant from sequence boundary ends: a_1 + a_2024 = a_5 + a_2020 = a_10 + a_2015 = dots$a_1 + a_{2024} = a_5 + a_{2020} = a_{10} + a_{2015} = \dots$ The sequence of inner indices follows an AP tracking loop: 5, 10, 15, dots, 2020$5, 10, 15, \dots, 2020$ Calculate internal block element count N$N$: 2020 = 5 + (N-1)5 implies 2015 = 5(N-1) implies N - 1 = 403 implies N = 404 text terms$2020 = 5 + (N-1)5 \implies 2015 = 5(N-1) \implies N - 1 = 403 \implies N = 404 \text{ terms}$ ### Step 1: Simplify Expression Equations Since the inner sequence contains 404 terms, they form exactly 202 symmetrical pairs. Adding a_1$a_1$ and a_2024$a_{2024}$ introduces one more pair, resulting in 203 identical sum blocks: 203(a_1 + a_2024) = 2233 implies a_1 + a_2024 = frac2233203 = 11$203(a_1 + a_{2024}) = 2233 \implies a_1 + a_{2024} = \frac{2233}{203} = 11$ ### Step 2: Evaluate the Total Sum Using the standard AP sum formula: S_2024 = frac20242(a_1 + a_2024) = 1012 times 11 = 11132$S_{2024} = \frac{2024}{2}(a_1 + a_{2024}) = 1012 \times 11 = 11132$ ### Pattern Recognition Progressions possess natural positional balance. Grouping symmetrical index pairs (a_k + a_n-k+1$a_k + a_{n-k+1}$) allows factoring out variable steps right away. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### 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