Let the coefficients of three consecutive terms T_r$T_{r}$, T_r+1$T_{r+1}$ and T_r+2$T_{r+2}$ in the binomial expansion of (a+b)^12$(a+b)^{12}$ be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (sqrt[4]3+sqrt[3]4)^12$(\sqrt[4]{3}+\sqrt[3]{4})^{12}$ . Then p+q$p+q$ is equal to :

### Related Formula (1 - 2x)^26 = sum_r=0^26 binom26r (-2x)^r$$(1 - 2x)^{26} = \sum_{r=0}^{26} \binom{26}{r} (-2x)^r$$ The coefficient extraction utilizes the distributive property over the polynomial (ax^2+bx+c)$(ax^2+bx+c)$. ### Core Logic Expansion of expression: (ax^2 + bx + c) sum_r=0^26 binom26r (-2x)^r$(ax^{2} + bx + c) \sum_{r=0}^{26} \binom{26}{r} (-2x)^{r}$ For the term in x^2$x^2$, we pick contributions yielding total power 2: a cdot (x^0 text term of sum) + b cdot (x^1 text term) + c cdot (x^2 text term) = 0$a \cdot (x^0 \text{ term of sum}) + b \cdot (x^1 \text{ term}) + c \cdot (x^2 \text{ term}) = 0$ a cdot binom260(-2)^0 + b cdot binom261(-2)^1 + c cdot binom262(-2)^2 = 0$a \cdot \binom{26}{0}(-2)^0 + b \cdot \binom{26}{1}(-2)^1 + c \cdot \binom{26}{2}(-2)^2 = 0$ a - 52b + 1300c = 0 dots(1)$a - 52b + 1300c = 0 \dots(1)$ ### Step 1: Formulate the system of linear equations For the term in x^3$x^3$ (set to 0): a cdot (x^1 text term) + b cdot (x^2 text term) + c cdot (x^3 text term) = 0$a \cdot (x^1 \text{ term}) + b \cdot (x^2 \text{ term}) + c \cdot (x^3 \text{ term}) = 0$ a cdot binom261(-2)^1 + b cdot binom262(-2)^2 + c cdot binom263(-2)^3 = 0$a \cdot \binom{26}{1}(-2)^1 + b \cdot \binom{26}{2}(-2)^2 + c \cdot \binom{26}{3}(-2)^3 = 0$ -52a + 1300b - 20800c = 0 dots(2)$-52a + 1300b - 20800c = 0 \dots(2)$ For the term in x^1$x^1$ (set to -56): b cdot (x^0 text term) + c cdot (x^1 text term) = -56$b \cdot (x^0 \text{ term}) + c \cdot (x^1 \text{ term}) = -56$ b cdot binom260(-2)^0 + c cdot binom261(-2)^1 = -56$b \cdot \binom{26}{0}(-2)^0 + c \cdot \binom{26}{1}(-2)^1 = -56$ b - 52c = -56 dots(3)$b - 52c = -56 \dots(3)$ ### Step 2: Solve the Linear System From (3), b = 52c - 56$b = 52c - 56$. Substitute into (2) divided by -52$-52$ to simplify: a - 25b + 400c = 0$a - 25b + 400c = 0$. Substitute into (1): a - 52b + 1300c = 0$a - 52b + 1300c = 0$. Subtracting: 27b - 900c = 0 Rightarrow b = frac1003c$27b - 900c = 0 \Rightarrow b = \frac{100}{3}c$. Wait, recalculating directly: From (2): -52a + 1300b - 20800c = 0 Rightarrow a - 25b + 400c = 0$-52a + 1300b - 20800c = 0 \Rightarrow a - 25b + 400c = 0$. From (1): a - 52b + 1300c = 0$a - 52b + 1300c = 0$. (a - 25b + 400c) - (a - 52b + 1300c) = 0 Rightarrow 27b - 900c = 0 Rightarrow 3b = 100c$ (a - 25b + 400c) - (a - 52b + 1300c) = 0 \Rightarrow 27b - 900c = 0 \Rightarrow 3b = 100c$. Substitute into (3): b - 52(3b/100) = -56 Rightarrow$b - 52(3b/100) = -56 \Rightarrow$ fractional results? Let's check coefficients accurately. binom263(-2)^3 = frac26 cdot 25 cdot 246 times (-8) = 2600 times -8 = -20800$\binom{26}{3}(-2)^3 = \frac{26 \cdot 25 \cdot 24}{6} \times (-8) = 2600 \times -8 = -20800$. Divide by -52: a - 25b + 400c = 0$a - 25b + 400c = 0$. Correct. (a - 52b + 1300c) - (a - 25b + 400c) = -27b + 900c = 0 Rightarrow 27b = 900c Rightarrow b = frac1003c$(a - 52b + 1300c) - (a - 25b + 400c) = -27b + 900c = 0 \Rightarrow 27b = 900c \Rightarrow b = \frac{100}{3}c$. This contradicts integer expectations. Let's substitute b$b$ from (3) directly. b = 52c - 56$b = 52c - 56$. 27(52c - 56) = 900c$27(52c - 56) = 900c$ 1404c - 1512 = 900c Rightarrow 504c = 1512 Rightarrow c = 3$1404c - 1512 = 900c \Rightarrow 504c = 1512 \Rightarrow c = 3$. Then b = 52(3) - 56 = 156 - 56 = 100$b = 52(3) - 56 = 156 - 56 = 100$. From (1): a - 52(100) + 1300(3) = 0 Rightarrow a - 5200 + 3900 = 0 Rightarrow a = 1300$a - 52(100) + 1300(3) = 0 \Rightarrow a - 5200 + 3900 = 0 \Rightarrow a = 1300$. ### Step 3: Compute final value We have a = 1300$a = 1300$, b = 100$b = 100$, c = 3$c = 3$. a + b + c = 1300 + 100 + 3 = 1403$$a + b + c = 1300 + 100 + 3 = 1403$$ ### Pattern Recognition Polynomial-Binomial product coefficient extractions strictly generate cascaded linear Diophantine-style equations. Solving from the lowest degree constraint (x^1$x^1$) upward sequentially unwinds the system cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

### Related Formula textSum of binomial coefficients: sum_k=0^n binomnk x^k = (1 + x)^n$$\text{Sum of binomial coefficients: } \sum_{k=0}^{n} \binom{n}{k} x^k = (1 + x)^n$$ textShifted binomial sum: sum_k=1^n binomnk x^k = (1+x)^n - 1$$\text{Shifted binomial sum: } \sum_{k=1}^{n} \binom{n}{k} x^k = (1+x)^n - 1$$ ### Core Logic We split the summation into two parts, express each as a binomial series expansion, and equate the resulting algebraic fraction to solve for alpha$\alpha$. ### Step 1: Split the summation The general term inside the summation can be written as: left( frac10^r+1 - 110^r right) = 10 - frac110^r = 10 - 10 left( frac110 right)^r+1$$\left( \frac{10^{r+1} - 1}{10^r} \right) = 10 - \frac{1}{10^r} = 10 - 10 \left( \frac{1}{10} \right)^{r+1}$$ Substitute this back into the sum: S = sum_r=0^10 left[ 10 - 10 left( frac110 right)^r+1 right] binom11r+1$$S = \sum_{r=0}^{10} \left[ 10 - 10 \left( \frac{1}{10} \right)^{r+1} \right] \binom{11}{r+1}$$ S = 10 sum_r=0^10 binom11r+1 - 10 sum_r=0^10 binom11r+1 left( frac110 right)^r+1$$S = 10 \sum_{r=0}^{10} \binom{11}{r+1} - 10 \sum_{r=0}^{10} \binom{11}{r+1} \left( \frac{1}{10} \right)^{r+1}$$ ### Step 2: Evaluate both parts of the sum For the first part, let s = r+1$s = r+1$: sum_r=0^10 binom11r+1 = sum_s=1^11 binom11s = 2^11 - 1$$\sum_{r=0}^{10} \binom{11}{r+1} = \sum_{s=1}^{11} \binom{11}{s} = 2^{11} - 1$$ For the second part, using s = r+1$s = r+1$: sum_r=0^10 binom11r+1 left( frac110 right)^r+1 = sum_s=1^11 binom11s left( frac110 right)^s = left( 1 + frac110 right)^11 - 1 = left( frac1110 right)^11 - 1$$\sum_{r=0}^{10} \binom{11}{r+1} \left( \frac{1}{10} \right)^{r+1} = \sum_{s=1}^{11} \binom{11}{s} \left( \frac{1}{10} \right)^s = \left( 1 + \frac{1}{10} \right)^{11} - 1 = \left( \frac{11}{10} \right)^{11} - 1$$ ### Step 3: Combine and find alpha Multiply both parts by 10: S = 10 left( 2^11 - 1 right) - 10 left[ left( frac1110 right)^11 - 1 right]$$S = 10 \left( 2^{11} - 1 \right) - 10 \left[ \left( \frac{11}{10} \right)^{11} - 1 \right]$$ S = 10 cdot 2^11 - 10 - 10 cdot frac11^1110^11 + 10 = 10 cdot 2^11 - frac11^1110^10$$S = 10 \cdot 2^{11} - 10 - 10 \cdot \frac{11^{11}}{10^{11}} + 10 = 10 \cdot 2^{11} - \frac{11^{11}}{10^{10}}$$ Express the first term with a denominator of 10^10$10^{10}$: 10 cdot 2^11 = frac10^11 cdot 2^1110^10 = frac20^1110^10$$10 \cdot 2^{11} = \frac{10^{11} \cdot 2^{11}}{10^{10}} = \frac{20^{11}}{10^{10}}$$ Thus, the total sum is: S = frac20^11 - 11^1110^10$$S = \frac{20^{11} - 11^{11}}{10^{10}}$$ Comparing this with fracalpha^11 - 11^1110^10$\frac{\alpha^{11} - 11^{11}}{10^{10}}$, we find: alpha = 20$\alpha = 20$ ### Pattern Recognition Binomial base scaling: Whenever you see a sum of the form sum a^r binomnr$\sum a^r \binom{n}{r}$, it is simply the expanded form of a shifted binomial expansion of (1 + a)^n$(1 + a)^n$. Factoring out scaling constants yields standard analytical forms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula Algebraic factorizations: x+1 = left(x^1/3right)^3 + 1^3 = left(x^1/3+1right)left(x^2/3-x^1/3+1right)$$x+1 = \left(x^{1/3}\right)^3 + 1^3 = \left(x^{1/3}+1\right)\left(x^{2/3}-x^{1/3}+1\right)$$ x-1 = left(sqrtxright)^2 - 1^2 = (sqrtx-1)(sqrtx+1)$$x-1 = \left(\sqrt{x}\right)^2 - 1^2 = (\sqrt{x}-1)(\sqrt{x}+1)$$ ### Core Logic Simplify inside the parentheses before using the Binomial general term formula. ### Step 1: Simplify Bracket Terms First fraction: fracx+1x^2/3-x^1/3+1 = x^1/3+1$$\frac{x+1}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$$ Second fraction: fracx-1x-x^1/2 = frac(sqrtx-1)(sqrtx+1)sqrtx(sqrtx-1) = fracsqrtx+1sqrtx = 1 + x^-1/2$$\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$$ Subtract the two simplified results: left(x^1/3+1right) - left(1+x^-1/2right) = x^1/3 - x^-1/2$$\left(x^{1/3}+1\right) - \left(1+x^{-1/2}\right) = x^{1/3} - x^{-1/2}$$ ### Step 2: Apply Binomial Theorem The expression simplifies to left(x^1/3 - x^-1/2right)^10$\left(x^{1/3} - x^{-1/2}\right)^{10}$. Write the general term T_r+1$T_{r+1}$: T_r+1 = binom10r left(x^1/3right)^10-r left(-x^-1/2right)^r = binom10r (-1)^r x^frac10-r3 - fracr2$$T_{r+1} = \binom{10}{r} \left(x^{1/3}\right)^{10-r} \left(-x^{-1/2}\right)^r = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$$ ### Step 3: Solve for Independent Term Set the net power of x$x$ to zero: frac10-r3 - fracr2 = 0 implies 20 - 2r - 3r = 0 implies 5r = 20 implies r = 4$$\frac{10-r}{3} - \frac{r}{2} = 0 \implies 20 - 2r - 3r = 0 \implies 5r = 20 \implies r = 4$$ Substitute r=4$r=4$ into the general term expression: textCoefficient = binom104 (-1)^4 = frac10 times 9 times 8 times 74 times 3 times 2 times 1 = 210$$\text{Coefficient} = \binom{10}{4} (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$ ### Pattern Recognition The initial fractions look complex but contain simple hidden identities (a^3+b^3$a^3+b^3$ and a^2-b^2$a^2-b^2$). Identifying these transforms a daunting fraction expansion into a classic two-term independent coefficient puzzle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula For trinomial expansion: (1+x+x^2)^n = sum_r=0^2n a_r x^r$$(1+x+x^2)^n = \sum_{r=0}^{2n} a_r x^r$$ Symmetric coefficients can be extracted by substituting x=1$x=1$ and x=-1$x=-1$. ### Core Logic Let's perform the substitution: - For x = 1$x = 1$: 3^10 = a_0 + a_1 + a_2 + dots + a_20 quad text--- (1)$$3^{10} = a_0 + a_1 + a_2 + \dots + a_{20} \quad \text{--- (1)}$$ - For x = -1$x = -1$: 1^10 = a_0 - a_1 + a_2 - a_3 + dots + a_20 quad text--- (2)$$1^{10} = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} \quad \text{--- (2)}$$ ### Step 1: Extracting odd term sum Subtracting (2) from (1): 3^10 - 1 = 2(a_1 + a_3 + a_5 + dots + a_19)$$3^{10} - 1 = 2(a_1 + a_3 + a_5 + \dots + a_{19})$$ a_1 + a_3 + a_5 + dots + a_19 = frac3^10 - 12$$a_1 + a_3 + a_5 + \dots + a_{19} = \frac{3^{10} - 1}{2}$$ Since 3^5 = 243 implies 3^10 = 59049$3^5 = 243 \implies 3^{10} = 59049$: a_1 + a_3 + a_5 + dots + a_19 = frac59049 - 12 = 29524$$a_1 + a_3 + a_5 + \dots + a_{19} = \frac{59049 - 1}{2} = 29524$$ ### Step 2: Finding a_2$a_2$ and k$k$ Using expansion formula to find a_2$a_2$ (coefficient of x^2$x^2$): (1 + x + x^2)^10 = sum frac10!p! q! r! x^q + 2r quad (p+q+r = 10)$$(1 + x + x^2)^{10} = \sum \frac{10!}{p! q! r!} x^{q + 2r} \quad (p+q+r = 10)$$ For q + 2r = 2$q + 2r = 2$: - Case 1: r=1, q=0 implies p=9$r=1, q=0 \implies p=9$. Coefficient = frac10!9! 0! 1! = 10$\frac{10!}{9! 0! 1!} = 10$ - Case 2: r=0, q=2 implies p=8$r=0, q=2 \implies p=8$. Coefficient = frac10!8! 2! 0! = 45$\frac{10!}{8! 2! 0!} = 45$ a_2 = 10 + 45 = 55$$a_2 = 10 + 45 = 55$$ Now substitute into the target relation: 29524 - 11(55) = 29524 - 605 = 28919$$29524 - 11(55) = 29524 - 605 = 28919$$ 121k = 28919 implies k = 239$$121k = 28919 \implies k = 239$$ ### Pattern Recognition Substituting standard complex roots or positive units is the fastest way to resolve sum of subset coefficients in polynomial expansions. Always use partition calculations for the early index terms (a_1, a_2$a_1, a_2$) rather than standard multinomial permutations to avoid calculation mistakes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula Binomial expansion for checking remainders: (1 + kx)^n = 1 + n(kx) + fracn(n-1)2(kx)^2 + dots equiv 1 pmod x$$(1 + kx)^n = 1 + n(kx) + \frac{n(n-1)}{2}(kx)^2 + \dots \equiv 1 \pmod x$$ ### Core Logic Let the target expression be N = left((64)^64right)^64$N = \left((64)^{64}\right)^{64}$. Using power rules, N = 64^64 times 64 = 64^64^2$N = 64^{64 \times 64} = 64^{64^2}$. Let the large exponent exponent be n = 64^2$n = 64^2$. We need to find the remainder of 64^n$64^n$ when divided by 7$7$. ### Step 1: Express Base in Terms of Modulus Observe that 64 = 63 + 1 = 7 times 9 + 1$64 = 63 + 1 = 7 \times 9 + 1$. Substituting this into the expression: N = (1 + 63)^n$N = (1 + 63)^n$ Expanding using the binomial theorem: N = 1 + ^nC_1(63) + ^nC_2(63)^2 + dots + ^nC_n(63)^n$$N = 1 + ^nC_1(63) + ^nC_2(63)^2 + \dots + ^nC_n(63)^n$$ N = 1 + 63 cdot lambda = 1 + 7(9lambda)$$N = 1 + 63 \cdot \lambda = 1 + 7(9\lambda)$$ Since 7(9lambda)$7(9\lambda)$ is perfectly divisible by 7$7$, the remaining term is 1$1$. ### Pattern Recognition Whenever the base can be written as km + 1$km + 1$, where m$m$ is the divisor, the value of (km + 1)^n equiv 1^n equiv 1 pmod m$(km + 1)^n \equiv 1^n \equiv 1 \pmod m$ instantly, regardless of the size or complexity of the exponent layout. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem