If sum_mathrmr = 0^10left(frac10^mathrmr + 1 - 110^mathrmrright) cdot binom11r + 1 = fracalpha^11 - 11^1110^10$\sum_{\mathrm{r} = 0}^{10}\left(\frac{10^{\mathrm{r} + 1} - 1}{10^{\mathrm{r}}}\right) \cdot \binom{11}{r + 1} = \frac{\alpha^{11} - 11^{11}}{10^{10}}$, then alpha$\alpha$ is equal to:

### Related Formula The coefficient of the r^textth$r^{\text{th}}$ term in the expansion (1+x)^m$(1+x)^m$ is written as binommr-1$\binom{m}{r-1}$. For three terms in A.P., their values satisfy: 2 cdot T_2 = T_1 + T_3$$2 \cdot T_2 = T_1 + T_3$$ ### Core Logic Let the total power exponent be m = n + 4$m = n + 4$. The coefficients of the 5^textth$5^{\text{th}}$, 6^textth$6^{\text{th}}$, and 7^textth$7^{\text{th}}$ terms are binomm4$\binom{m}{4}$, binomm5$\binom{m}{5}$, and binomm6$\binom{m}{6}$ respectively. Since they form an arithmetic progression: 2 cdot binomm5 = binomm4 + binomm6$$2 \cdot \binom{m}{5} = \binom{m}{4} + \binom{m}{6}$$ Rearrange using the recurrence addition identity properties: 4 cdot binomm5 = left[ binomm4 + binomm5 right] + left[ binomm5 + binomm6 right]$$4 \cdot \binom{m}{5} = \left[ \binom{m}{4} + \binom{m}{5} \right] + \left[ \binom{m}{5} + \binom{m}{6} \right]$$ 4 cdot binomm5 = binomm+15 + binomm+16$$4 \cdot \binom{m}{5} = \binom{m+1}{5} + \binom{m+1}{6}$$ 4 cdot binomm5 = binomm+26$$4 \cdot \binom{m}{5} = \binom{m+2}{6}$$ ### Step 1: Solve the Combinatorial Fraction for m$m$ Expand the combinations using factorials: 4 cdot fracm!5!(m-4)! = frac(m+2)!6!(m-4)!$$4 \cdot \frac{m!}{5!(m-4)!} = \frac{(m+2)!}{6!(m-4)!}$$ Cancel out (m-4)!$(m-4)!$ from both denominators: 4 cdot fracm!120 = frac(m+2)(m+1)m!720$$4 \cdot \frac{m!}{120} = \frac{(m+2)(m+1)m!}{720}$$ 4 = frac(m+2)(m+1)6$$4 = \frac{(m+2)(m+1)}{6}$$ 24 = m^2 + 3m + 2 implies m^2 + 3m - 22 = 0$$24 = m^2 + 3m + 2 \implies m^2 + 3m - 22 = 0$$ Wait, let's re-verify the step using the direct ratio computation: 2 = fracbinomm4binomm5 + fracbinomm6binomm5 = frac5m-4 + fracm-56$$2 = \frac{\binom{m}{4}}{\binom{m}{5}} + \frac{\binom{m}{6}}{\binom{m}{5}} = \frac{5}{m-4} + \frac{m-5}{6}$$ 2 = frac30 + (m-4)(m-5)6(m-4)$$2 = \frac{30 + (m-4)(m-5)}{6(m-4)}$$ 12(m-4) = 30 + m^2 - 9m + 20$$12(m-4) = 30 + m^2 - 9m + 20$$ 12m - 48 = m^2 - 9m + 50 implies m^2 - 21m + 98 = 0$$12m - 48 = m^2 - 9m + 50 \implies m^2 - 21m + 98 = 0$$ Factor the quadratic equation: (m-7)(m-14) = 0 implies m = 7 text or m = 14$$(m-7)(m-14) = 0 \implies m = 7 \text{ or } m = 14$$ ### Step 2: Connect back to the problem constraints Since m = n + 4$m = n + 4$: - If m = 14 implies n + 4 = 14 implies n = 10$m = 14 \implies n + 4 = 14 \implies n = 10$ (this is rejected because the problem states n neq 10$n \neq 10$). - If m = 7 implies n + 4 = 7 implies n = 3$m = 7 \implies n + 4 = 7 \implies n = 3$ (this value is accepted). Thus, the binomial expansion exponent is exactly m = 7$m = 7$. ### Step 3: Extract Maximum Binomial Coefficient For an odd exponent power m = 7$m = 7$, the maximum binomial coefficient corresponds to the middle terms: textMax Coefficient = binom73 = binom74 = frac7 cdot 6 cdot 53 cdot 2 cdot 1 = 35$$\text{Max Coefficient} = \binom{7}{3} = \binom{7}{4} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$$ ### Pattern Recognition For consecutive binomial coefficients binomnr-1, binomnr, binomnr+1$\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}$ in A.P., the power parameters satisfy the standard identity (n-2r)^2 = n+2$(n-2r)^2 = n+2$, which allows quick calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula General term of binomial expansion (x + y)^n$(x + y)^n$: T_k+1 = binomnk x^n-k y^k$$T_{k+1} = \binom{n}{k} x^{n-k} y^k$$ Condition for three terms A, B, C$A, B, C$ to be in G.P.: B^2 = A cdot C$B^2 = A \cdot C$ ### Core Logic Part 1: Coefficients of T_r, T_r+1, T_r+2$T_r, T_{r+1}, T_{r+2}$ are binom12r-1, binom12r, binom12r+1$\binom{12}{r-1}, \binom{12}{r}, \binom{12}{r+1}$. Since they are in G.P.: left[binom12rright]^2 = binom12r-1 cdot binom12r+1$$\left[\binom{12}{r}\right]^2 = \binom{12}{r-1} \cdot \binom{12}{r+1}$$ fracbinom12rbinom12r-1 = fracbinom12r+1binom12r implies frac12-r+1r = frac12-rr+1$$\frac{\binom{12}{r}}{\binom{12}{r-1}} = \frac{\binom{12}{r+1}}{\binom{12}{r}} \implies \frac{12-r+1}{r} = \frac{12-r}{r+1}$$ (13-r)(r+1) = r(12-r) implies 13r + 13 - r^2 - r = 12r - r^2$$(13-r)(r+1) = r(12-r) \implies 13r + 13 - r^2 - r = 12r - r^2$$ 12r + 13 = 12r implies 13 = 0 quad (textNot possible)$$12r + 13 = 12r \implies 13 = 0 \quad (\text{Not possible})$$ Thus, no real integer solution for r$r$ exists, so p = 0$p = 0$. ### Step 1: Calculate Rational Terms Sum q Part 2: Rational terms in expansion of (3^1/4 + 4^1/3)^12$(3^{1/4} + 4^{1/3})^{12}$. General term: T_k+1 = binom12k (3^1/4)^12-k (4^1/3)^k = binom12k 3^frac12-k4 4^frack3$$T_{k+1} = \binom{12}{k} (3^{1/4})^{12-k} (4^{1/3})^k = \binom{12}{k} 3^{\frac{12-k}{4}} 4^{\frac{k}{3}}$$ For the term to be rational, frac12-k4$\frac{12-k}{4}$ and \frac{k}{3} must both be integers: - k$k$ must be a multiple of 3: k in \0, 3, 6, 9, 12\$k \in \{0, 3, 6, 9, 12\}$ - 12-k$12-k$ must be a multiple of 4, so k$k$ must be a multiple of 4: k in \0, 4, 8, 12\$k \in \{0, 4, 8, 12\}$ The common values for k$k$ are k = 0$k = 0$ and k = 12$k = 12$. - At k = 0$k = 0$: T_1 = binom120 3^3 4^0 = 1 times 27 times 1 = 27$$T_1 = \binom{12}{0} 3^3 4^0 = 1 \times 27 \times 1 = 27$$ - At k = 12$k = 12$: T_13 = binom1212 3^0 4^4 = 1 times 1 times 256 = 256$$T_{13} = \binom{12}{12} 3^0 4^4 = 1 \times 1 \times 256 = 256$$ Sum of rational terms q = 27 + 256 = 283$q = 27 + 256 = 283$. ### Step 2: Final Combination p + q = 0 + 283 = 283$$p + q = 0 + 283 = 283$$ ### Pattern Recognition To find common values for divisibility constraints, look for multiples of the least common multiple textlcm(3, 4) = 12$\text{\lcm}(3, 4) = 12$ within the range [0, 12]$[0, 12]$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula T_r+1 = binomnr a^n-r b^r$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ ### Core Logic The general term in the expansion is: T_r+1 = binomnr (7)^fracn-r3 (11)^fracr12$$T_{r+1} = \binom{n}{r} (7)^{\frac{n-r}{3}} (11)^{\frac{r}{12}}$$ For the term to be integral, both powers fracn-r3$\frac{n-r}{3}$ and fracr12$\frac{r}{12}$ must be integers. This dictates that r$r$ must be a multiple of 12$12$ (r = 12k$r = 12k$, where k$k$ is a non-negative integer). ### Step 1: Setup total count equation The values r$r$ can take are 0, 12, 24, dots, 12k$0, 12, 24, \dots, 12k$. The total number of terms is given as 183$183$. Since counting starts from k=0$k=0$, the maximum value of k$k$ is: k_textmax = 183 - 1 = 182$$k_{\text{\max}} = 183 - 1 = 182$$ ### Step 2: Calculate minimum n The maximum index value r$r$ required to achieve this count is: r_textmax = 12 times 182 = 2184$$r_{\text{\max}} = 12 \times 182 = 2184$$ Hence, the least value of n$n$ must be 2184$2184$ to encompass all 183 integral terms. ### Pattern Recognition Number of terms formula when tracking steps of size L$L$: textTotal Terms = lfloor fracnL rfloor + 1$\text{Total Terms} = \lfloor \frac{n}{L} \rfloor + 1$. Isolate n$n$ directly to compute bounds rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula General term in a binomial expansion (1+t)^n$(1+t)^n$ is given by: T_r+1 = binomnr t^r$$T_{r+1} = \binom{n}{r} t^r$$ ### Core Logic Let's simplify the algebraic structure of the product expression first: left(1+frac1xright)^6(1+x^2)^7(1-x^3)^8 = frac(x+1)^6 (1+x^2)^7 (1-x^3)^8x^6$$\left(1+\frac{1}{x}\right)^{6}(1+x^{2})^{7}(1-x^{3})^{8} = \frac{(x+1)^6 (1+x^2)^7 (1-x^3)^8}{x^6}$$ Finding the coefficient of x^30$x^{30}$ in this full product is equivalent to finding the coefficient of x^36$x^{36}$ in the numerator expansion: textTarget = textCoefficient of x^36 text in (1+x)^6 (1+x^2)^7 (1-x^3)^8$$\text{Target} = \text{Coefficient of } x^{36} \text{ in } (1+x)^6 (1+x^2)^7 (1-x^3)^8$$ ### Step 1: Setting up General Term Constraints The product of the three general terms is: binom6r_1 x^r_1 cdot binom7r_2 (x^2)^r_2 cdot binom8r_3 (-x^3)^r_3 = binom6r_1binom7r_2binom8r_3 (-1)^r_3 x^r_1 + 2r_2 + 3r_3$$\binom{6}{r_1} x^{r_1} \cdot \binom{7}{r_2} (x^2)^{r_2} \cdot \binom{8}{r_3} (-x^3)^{r_3} = \binom{6}{r_1}\binom{7}{r_2}\binom{8}{r_3} (-1)^{r_3} x^{r_1 + 2r_2 + 3r_3}$$ We require the total exponent to equal 36$36$: r_1 + 2r_2 + 3r_3 = 36$$r_1 + 2r_2 + 3r_3 = 36$$ with boundaries 0 le r_1 le 6$0 \le r_1 \le 6$, 0 le r_2 le 7$0 \le r_2 \le 7$, 0 le r_3 le 8$0 \le r_3 \le 8$. ### Step 2: Case Analysis by r3 Let's evaluate non-vanishing integer combinations case-by-case: - **Case I: r_3 = 8$r_3 = 8$** r_1 + 2r_2 = 36 - 24 = 12$$r_1 + 2r_2 = 36 - 24 = 12$$ - r_2 = 6, r_1 = 0 implies binom60binom76binom88(-1)^8 = 1 times 7 times 1 = 7$r_2 = 6, r_1 = 0 \implies \binom{6}{0}\binom{7}{6}\binom{8}{8}(-1)^8 = 1 \times 7 \times 1 = 7$ - r_2 = 5, r_1 = 2 implies binom62binom75binom88(-1)^8 = 15 times 21 times 1 = 315$r_2 = 5, r_1 = 2 \implies \binom{6}{2}\binom{7}{5}\binom{8}{8}(-1)^8 = 15 \times 21 \times 1 = 315$ - r_2 = 4, r_1 = 4 implies binom64binom74binom88(-1)^8 = 15 times 35 times 1 = 525$r_2 = 4, r_1 = 4 \implies \binom{6}{4}\binom{7}{4}\binom{8}{8}(-1)^8 = 15 \times 35 \times 1 = 525$ - r_2 = 3, r_1 = 6 implies binom66binom73binom88(-1)^8 = 1 times 35 times 1 = 35$r_2 = 3, r_1 = 6 \implies \binom{6}{6}\binom{7}{3}\binom{8}{8}(-1)^8 = 1 \times 35 \times 1 = 35$ - **Case II: r_3 = 7$r_3 = 7$** r_1 + 2r_2 = 36 - 21 = 15$$r_1 + 2r_2 = 36 - 21 = 15$$ - r_2 = 7, r_1 = 1 implies binom61binom77binom87(-1)^7 = 6 times 1 times 8 times (-1) = -48$r_2 = 7, r_1 = 1 \implies \binom{6}{1}\binom{7}{7}\binom{8}{7}(-1)^7 = 6 \times 1 \times 8 \times (-1) = -48$ - r_2 = 6, r_1 = 3 implies binom63binom76binom87(-1)^7 = 20 times 7 times 8 times (-1) = -1120$r_2 = 6, r_1 = 3 \implies \binom{6}{3}\binom{7}{6}\binom{8}{7}(-1)^7 = 20 \times 7 \times 8 \times (-1) = -1120$ - r_2 = 5, r_1 = 5 implies binom65binom75binom87(-1)^7 = 6 times 21 times 8 times (-1) = -1008$r_2 = 5, r_1 = 5 \implies \binom{6}{5}\binom{7}{5}\binom{8}{7}(-1)^7 = 6 \times 21 \times 8 \times (-1) = -1008$ - **Case III: r_3 = 6$r_3 = 6$** r_1 + 2r_2 = 36 - 18 = 18$$r_1 + 2r_2 = 36 - 18 = 18$$ - r_2 = 7, r_1 = 4 implies binom64binom77binom86(-1)^6 = 15 times 1 times 28 = 420$r_2 = 7, r_1 = 4 \implies \binom{6}{4}\binom{7}{7}\binom{8}{6}(-1)^6 = 15 \times 1 \times 28 = 420$ - r_2 = 6, r_1 = 6 implies binom66binom76binom86(-1)^6 = 1 times 7 times 28 = 196$r_2 = 6, r_1 = 6 \implies \binom{6}{6}\binom{7}{6}\binom{8}{6}(-1)^6 = 1 \times 7 \times 28 = 196$ ### Step 3: Summation for Alpha Summing all calculated values: alpha = (7 + 315 + 525 + 35) + (-48 - 1120 - 1008) + (420 + 196)$$\alpha = (7 + 315 + 525 + 35) + (-48 - 1120 - 1008) + (420 + 196)$$ alpha = 882 - 2176 + 616 = -678$$\alpha = 882 - 2176 + 616 = -678$$ Thus, the absolute value is: |alpha| = 678$|\alpha| = 678$ ### Pattern Recognition Sees: Multi-product polynomial coefficient extraction problem. Trap: Remember that the negative sign inside (1-x^3)^8$(1-x^3)^8$ alters the polarity of terms based on whether r_3$r_3$ is odd or even. Always track the (-1)^r_3$(-1)^{r_3}$ factor carefully. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula (9k + 1)^n equiv 1 pmod 9$$(9k + 1)^n \equiv 1 \pmod 9$$ ### Core Logic Let the giant power tower expression be variable parameter t = 32^32$t = 32^{32}$. Re-expressing baseline number base 64: 64 = 8^2$64 = 8^2$ So, the total term evaluates to: 64^t = (8^2)^t = 8^2t$$64^t = (8^2)^t = 8^{2t}$$ ### Step 1: Evaluating Modulo Properties We can re-express baseline 8 as (9 - 1)$(9 - 1)$: 8^2t = (9 - 1)^2t$$8^{2t} = (9 - 1)^{2t}$$ Expanding this expression using the Binomial Theorem: (9 - 1)^2t = binom2t09^2t - dots + binom2t2t(-1)^2t$$(9 - 1)^{2t} = \binom{2t}{0}9^{2t} - \dots + \binom{2t}{2t}(-1)^{2t}$$ Since the exponent 2t$2t$ is an even number, (-1)^2t = 1$(-1)^{2t} = 1$. Therefore: (9 - 1)^2t = 9k + 1 equiv 1 pmod 9$$(9 - 1)^{2t} = 9k + 1 \equiv 1 \pmod 9$$ Thus, the remainder obtained when divided by 9 is exactly 1. ### Pattern Recognition When evaluating exponents modulo M$M$, reduce the base first. Since 64 equiv 1 pmod 9$64 \equiv 1 \pmod 9$, any integer power of 64 trivially leaves a remainder of 1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem