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 T_r+1 = binomnr a^n-r b^r$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ ### Core Logic General term in the expansion of left(7^1/2 + 11^1/6right)^824$\left(7^{1/2} + 11^{1/6}\right)^{824}$ is: t_r+1 = binom824r (7)^frac824-r2 (11)^r/6$$t_{r+1} = \binom{824}{r} (7)^{\frac{824-r}{2}} (11)^{r/6}$$ For the term to be integral, both powers must be integers. This means: 1) frac824 - r2$\frac{824 - r}{2}$ must be an integer, which means r$r$ must be even. 2) fracr6$\frac{r}{6}$ must be an integer, which means r$r$ must be a multiple of 6. Since any multiple of 6 is already even, the condition reduces to: r$r$ must be a multiple of 6. ### Step 1: Finding valid values of r The possible values for r$r$ are 0, 1, 2, dots, 824$0, 1, 2, \dots, 824$. Valid r = 0, 6, 12, dots, 822$r = 0, 6, 12, \dots, 822$. This forms an arithmetic progression with first term a=0$a=0$, common difference d=6$d=6$, and last term L=822$L=822$. L = a + (n-1)d$L = a + (n-1)d$ 822 = 0 + (n-1)6$$822 = 0 + (n-1)6$$ n - 1 = frac8226 = 137$$n - 1 = \frac{822}{6} = 137$$ n = 138$n = 138$ Thus, there are 138 integral terms. ### Pattern Recognition Finding rational/integral terms in a binomial expansion strictly requires finding the LCM of the fractional power denominators, then counting multiples up to n$n$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

### Related Formula textSum of coefficients in a polynomial P(x) text is found by putting x=1.$$\text{Sum of coefficients in a polynomial } P(x) \text{ is found by putting } x=1.$$ textSum of G.P.: S_n = fraca(r^n - 1)r - 1$$\text{Sum of G.P.: } S_n = \frac{a(r^n - 1)}{r - 1}$$ ### Core Logic Let the expanded polynomial be P(x)$P(x)$. The sum of its coefficients is sum alpha_r = P(1)$\sum \alpha_r = P(1)$. Substitute x = 1$x = 1$ into the given expression: P(1) = 4^n-1 + 4^n-2(3) + 4^n-3(3^2) + dots + 3^n-1$$P(1) = 4^{n-1} + 4^{n-2}(3) + 4^{n-3}(3^2) + \dots + 3^{n-1}$$ This is a Geometric Progression with first term a = 4^n-1$a = 4^{n-1}$ and common ratio r = 3/4$r = 3/4$. There are n$n$ terms. P(1) = 4^n-1 frac1 - (3/4)^n1 - 3/4 = 4^n-1 frac1 - (3/4)^n1/4 = 4^n left(1 - frac3^n4^nright) = 4^n - 3^n$$P(1) = 4^{n-1} \frac{1 - (3/4)^n}{1 - 3/4} = 4^{n-1} \frac{1 - (3/4)^n}{1/4} = 4^n \left(1 - \frac{3^n}{4^n}\right) = 4^n - 3^n$$ Comparing this with beta^n - gamma^n$\beta^n - \gamma^n$, we get: beta = 4, quad gamma = 3$$\beta = 4, \quad \gamma = 3$$ Calculate beta^2 + gamma^2$\beta^2 + \gamma^2$: beta^2 + gamma^2 = 4^2 + 3^2 = 16 + 9 = 25$$\beta^2 + \gamma^2 = 4^2 + 3^2 = 16 + 9 = 25$$ ### Pattern Recognition Substituting x=1$x=1$ immediately bypasses expanding individual x^r$x^r$ terms for questions asking for sum of coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem Class 11 Maths: Sequences and Series

### Core Logic To find the sum of all coefficients in a polynomial expansion, substitute x = 1$x = 1$. a = (1 - 2(1) + 2(1)^2)^2023 (3 - 4(1)^2 + 2(1)^3)^2024$$a = (1 - 2(1) + 2(1)^2)^{2023} (3 - 4(1)^2 + 2(1)^3)^{2024}$$ a = (1)^2023 (1)^2024 = 1$$a = (1)^{2023} (1)^{2024} = 1$$ ### Step 1: Evaluate Limit for b Evaluate b = lim_x to 0 fracint_0^x fracln(1 + t)1 + t^2024 dtx^2$b = \lim_{x \to 0} \frac{\int_{0}^{x} \frac{\ln(1 + t)}{1 + t^{2024}} dt}{x^2}$ Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz): b = lim_x to 0 fracfracln(1 + x)1 + x^20242x = lim_x to 0 fracln(1 + x)x times frac12(1 + x^2024)$$b = \lim_{x \to 0} \frac{\frac{\ln(1 + x)}{1 + x^{2024}}}{2x} = \lim_{x \to 0} \frac{\ln(1 + x)}{x} \times \frac{1}{2(1 + x^{2024})}$$ b = 1 times frac12 = frac12$$b = 1 \times \frac{1}{2} = \frac{1}{2}$$ ### Step 2: Analyze Common Roots The given second equation is 2bx^2 + ax + 4 = 0$2bx^2 + ax + 4 = 0$. Substitute a = 1$a = 1$ and b = frac12$b = \frac{1}{2}$: 2left(frac12right)x^2 + 1(x) + 4 = 0 implies x^2 + x + 4 = 0$$2\left(\frac{1}{2}\right)x^2 + 1(x) + 4 = 0 \implies x^2 + x + 4 = 0$$ The discriminant of x^2 + x + 4 = 0$x^2 + x + 4 = 0$ is D = 1 - 16 < 0$D = 1 - 16 < 0$. Roots are non-real complex conjugates. ### Step 3: Final Ratio Since c, d, e in mathbbR$c, d, e \in \mathbb{R}$ and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients must be proportional: fracc1 = fracd1 = frace4$$\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$$ This implies d : c : e = 1 : 1 : 4$d : c : e = 1 : 1 : 4$. ### Pattern Recognition If a quadratic equation with real coefficients shares a common root with another quadratic having complex roots (D < 0$D < 0$), both roots must be shared, meaning their coefficients are directly proportional. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem Class 12 Maths: Limits and Derivatives Class 11 Maths: Quadratic Equations

### Core Logic (1+x)(1-x^2) left( left(1 + frac1xright)^3 right)^5$$(1+x)(1-x^2) \left( \left(1 + \frac{1}{x}\right)^3 \right)^5$$ = (1+x)(1-x)(1+x) frac(x+1)^15x^15$$= (1+x)(1-x)(1+x) \frac{(x+1)^{15}}{x^{15}}$$ = frac(1-x)(1+x)^17x^15$$= \frac{(1-x)(1+x)^{17}}{x^{15}}$$ = frac(1+x)^17 - x(1+x)^17x^15$$= \frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$$ ### Step 1: Find Coefficient of x^3 To find coeff of x^3$x^3$ in frac(1+x)^17 - x(1+x)^17x^15$\frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$, we need the coeff of x^18$x^{18}$ in the numerator (1+x)^17 - x(1+x)^17$(1+x)^{17} - x(1+x)^{17}$. The maximum power of x$x$ in (1+x)^17$(1+x)^{17}$ is 17$17$, and in x(1+x)^17$x(1+x)^{17}$ is 18$18$. Coeff of x^18$x^{18}$ in (1+x)^17$(1+x)^{17}$ is 0. Coeff of x^18$x^{18}$ in x(1+x)^17$x(1+x)^{17}$ is the coeff of x^17$x^{17}$ in (1+x)^17$(1+x)^{17}$, which is binom1717 = 1$\binom{17}{17} = 1$. Thus, coeff of x^18$x^{18}$ in the numerator is 0 - 1 = -1$0 - 1 = -1$. ### Step 2: Find Coefficient of x^{-13} To find coeff of x^-13$x^{-13}$, we need the coeff of x^2$x^2$ in the numerator (1+x)^17 - x(1+x)^17$(1+x)^{17} - x(1+x)^{17}$. Coeff of x^2$x^2$ in (1+x)^17$(1+x)^{17}$ is binom172$\binom{17}{2}$. Coeff of x^2$x^2$ in x(1+x)^17$x(1+x)^{17}$ is coeff of x^1$x^1$ in (1+x)^17$(1+x)^{17}$, which is binom171$\binom{17}{1}$. Value = binom172 - binom171 = frac17 times 162 - 17 = 136 - 17 = 119$\binom{17}{2} - \binom{17}{1} = \frac{17 \times 16}{2} - 17 = 136 - 17 = 119$. ### Step 3: Final Sum Sum of coefficients = -1 + 119 = 118$-1 + 119 = 118$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem