### Related Formula (10k - 1)^n = dots + binomnn-1(10k)(-1)^n-1 + (-1)^n$(10k - 1)^n = \dots + \binom{n}{n-1}(10k)(-1)^{n-1} + (-1)^n$ ### Core Logic Break the base parameter into a multiple of 10 form (1920 - 1$1920 - 1$) and use binomial expansions to decouple high factor blocks from final terminal digits. ### Step 1: Set Up Binomial Expansion (1919)^1919 = (1920 - 1)^1919$(1919)^{1919} = (1920 - 1)^{1919}$ = binom19190(1920)^1919 - dots + binom19191918(1920)^1 - binom19191919$= \binom{1919}{0}(1920)^{1919} - \dots + \binom{1919}{1918}(1920)^1 - \binom{1919}{1919}$ ### Step 2: Isolate Low Factor Coefficients All higher structural tracks sitting above index points present factors multiple loops over 100. Isolate trailing values: = 100lambda + 1919 times 1920 - 1$= 100\lambda + 1919 \times 1920 - 1$ ### Step 3: Deduce Final Trailing Quotient Product 1919 times 1920 - 1 = 3684480 - 1 = 3684479$1919 \times 1920 - 1 = 3684480 - 1 = 3684479$ Trailing structural indicators: 79$79$. Product calculation response: 7 times 9 = 63$7 \times 9 = 63$ ### Pattern Recognition Using expansions around tens bases shifts focus entirely to the last two expansion expressions to find digit answers rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula Modular arithmetic congruence rules: A equiv B pmodm implies A^n equiv B^n pmodm$A \equiv B \pmod{m} \implies A^n \equiv B^n \pmod{m}$ ### Core Logic We need to compute 7^103 pmod23$7^{103} \pmod{23}$. Let's break the exponent down into manageable powers: 7^103 = 7 cdot (7^2)^51 = 7 cdot (49)^51$7^{103} = 7 \cdot (7^2)^{51} = 7 \cdot (49)^{51}$ Since 49 equiv 3 pmod23$49 \equiv 3 \pmod{23}$: 7^103 equiv 7 cdot 3^51 pmod23$7^{103} \equiv 7 \cdot 3^{51} \pmod{23}$ ### Step 1: Simplify Higher Power Components Break down 3^51$3^{51}$ using 3^3 = 27 equiv 4 pmod23$3^3 = 27 \equiv 4 \pmod{23}$: 3^51 = (3^3)^17 = 27^17 equiv 4^17 pmod23$3^{51} = (3^3)^{17} = 27^{17} \equiv 4^{17} \pmod{23}$ So the expression becomes: 7 cdot 4^17 = 7 cdot 4 cdot (4^2)^8 = 28 cdot 16^8$7 \cdot 4^{17} = 7 \cdot 4 \cdot (4^2)^8 = 28 \cdot 16^8$ Since 28 equiv 5 pmod23$28 \equiv 5 \pmod{23}$ and 16 equiv -7 pmod23$16 \equiv -7 \pmod{23}$: equiv 5 cdot (-7)^8 = 5 cdot 7^8 pmod23$\equiv 5 \cdot (-7)^8 = 5 \cdot 7^8 \pmod{23}$ ### Step 2: Final Remainder Evaluation Now compute 7^8$7^8$ using 7^2 = 49 equiv 3 pmod23$7^2 = 49 \equiv 3 \pmod{23}$: 7^8 = (7^2)^4 equiv 3^4 = 81 pmod23$7^8 = (7^2)^4 \equiv 3^4 = 81 \pmod{23}$ Since 81 equiv 12 pmod23$81 \equiv 12 \pmod{23}$: textExpression equiv 5 cdot 12 = 60 pmod23$\text{Expression} \equiv 5 \cdot 12 = 60 \pmod{23}$ Dividing 60 by 23 (23 times 2 = 46$23 \times 2 = 46$) yields a remainder of: 60 - 46 = 14$60 - 46 = 14$ ### Pattern Recognition Using negative remainders (like 16 equiv -7$16 \equiv -7$) reduces large multiplication outputs instantly during binary power modular loops. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula General term in the binomial expansion of (a+b)^n$(a+b)^n$: T_r+1 = ^nC_r cdot a^n-r cdot b^r$T_{r+1} = {^nC_r} \cdot a^{n-r} \cdot b^r$ ### Core Logic For a term to be rational in (2+sqrt3)^8$(2+\sqrt{3})^8$, the power of sqrt3$\sqrt{3}$ must be an even integer[cite: 1218]. Thus, r$r$ can only take even values from 0$0$ to 8$8$ [cite: 1219, 1220]: r in \0, 2, 4, 6, 8\$r \in \{0, 2, 4, 6, 8\}$ [cite: 1219, 1220] ### Step 1: Calculating individual rational terms Sum up terms explicitly for all valid values of r$r$ [cite: 1219, 1220]: textSum = ^8C_0(2)^8 + ^8C_2(2)^6(sqrt3)^2 + ^8C_4(2)^4(sqrt3)^4 + ^8C_6(2)^2(sqrt3)^6 + ^8C_8(sqrt3)^8$\text{Sum} = {^8C_0}(2)^8 + {^8C_2}(2)^6(\sqrt{3})^2 + {^8C_4}(2)^4(\sqrt{3})^4 + {^8C_6}(2)^2(\sqrt{3})^6 + {^8C_8}(\sqrt{3})^8$ [cite: 1219, 1220] Compute the individual numeric values [cite: 1221]: - r=0: 1 cdot 256 = 256$r=0: 1 \cdot 256 = 256$ - r=2: 28 cdot 64 cdot 3 = 5376$r=2: 28 \cdot 64 \cdot 3 = 5376$ - r=4: 70 cdot 16 cdot 9 = 10080$r=4: 70 \cdot 16 \cdot 9 = 10080$ - r=28 cdot 4 cdot 27 = 3024$r=28 \cdot 4 \cdot 27 = 3024$ - r=8: 1 cdot 1 cdot 81 = 81$r=8: 1 \cdot 1 \cdot 81 = 81$ textTotal Sum = 256 + 5376 + 10080 + 3024 + 81 = 18817$\text{Total Sum} = 256 + 5376 + 10080 + 3024 + 81 = 18817$ [cite: 1221, 1222] ### Pattern Recognition Notice that the sum of all rational terms can also be viewed as the rational part of the expanded configuration, equivalent to frac(2+sqrt3)^8 + (2-sqrt3)^82$\frac{(2+\sqrt{3})^8 + (2-\sqrt{3})^8}{2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula 1. sum_r=1^n r cdot ^nC_r x^r = nx(1+x)^n-1$\sum_{r=1}^n r \cdot {^nC_r} x^r = nx(1+x)^{n-1}$ 2. Binomial Theorem expansion: sum_r=0^n ^nC_r x^r = (1+x)^n$\sum_{r=0}^n {^nC_r} x^r = (1+x)^n$ ### Core Logic Split the given series summation into two independent parts [cite: 1316]: textSum = sum_r=1^9 fracr2^r cdot ^9C_r + 3sum_r=1^9 frac12^r cdot ^9C_r$\text{Sum} = \sum_{r=1}^9 \frac{r}{2^r} \cdot {^9C_r} + 3\sum_{r=1}^9 \frac{1}{2^r} \cdot {^9C_r}$ [cite: 1316] Simplify the first sub-sum using the index relation r cdot ^9C_r = 9 cdot ^8C_r-1$r \cdot {^9C_r} = 9 \cdot {^8C_{r-1}}$ [cite: 1316]: sum_r=1^9 frac92^r cdot ^8C_r-1 = frac92sum_r=1^9 ^8C_r-1left(frac12right)^r-1 = frac92left(1 + frac12right)^8 = frac92left(frac32right)^8$\sum_{r=1}^9 \frac{9}{2^r} \cdot {^8C_{r-1}} = \frac{9}{2}\sum_{r=1}^9 {^8C_{r-1}}\left(\frac{1}{2}\right)^{r-1} = \frac{9}{2}\left(1 + \frac{1}{2}\right)^8 = \frac{9}{2}\left(\frac{3}{2}\right)^8$ [cite: 1316] Simplify the second sub-sum by including missing index r=0$r=0$ [cite: 1316]: 3left[sum_r=0^9 ^9C_rleft(frac12right)^r - 1right] = 3left[left(1 + frac12right)^9 - 1right] = 3left(frac32right)^9 - 3$3\left[\sum_{r=0}^9 {^9C_r}\left(\frac{1}{2}\right)^r - 1\right] = 3\left[\left(1 + \frac{1}{2}\right)^9 - 1\right] = 3\left(\frac{3}{2}\right)^9 - 3$ [cite: 1316] ### Step 1: Combining the components Combine both evaluations to fit into requested representation shape [cite: 1316]: textTotal Sum = frac92left(frac32right)^8 + 3left(frac32right)^9 - 3$\text{Total Sum} = \frac{9}{2}\left(\frac{3}{2}\right)^8 + 3\left(\frac{3}{2}\right)^9 - 3$ [cite: 1316] Convert the fractional leading term [cite: 1316]: frac92left(frac32right)^8 = 3 cdot frac32left(frac32right)^8 = 3left(frac32right)^9$\frac{9}{2}\left(\frac{3}{2}\right)^8 = 3 \cdot \frac{3}{2}\left(\frac{3}{2}\right)^8 = 3\left(\frac{3}{2}\right)^9$ [cite: 1316] textTotal Sum = 3left(frac32right)^9 + 3left(frac32right)^9 - 3 = 6left(frac32right)^9 - 3$\text{Total Sum} = 3\left(\frac{3}{2}\right)^9 + 3\left(\frac{3}{2}\right)^9 - 3 = 6\left(\frac{3}{2}\right)^9 - 3$ [cite: 1316] Matching coefficients gives [cite: 1317]: alpha = 6, quad beta = 3$\alpha = 6, \quad \beta = 3$ [cite: 1317] Evaluate the required squared value [cite: 1317]: (alpha + beta)^2 = (6 + 3)^2 = 81$(\alpha + \beta)^2 = (6 + 3)^2 = 81$ [cite: 1317] ### Pattern Recognition Splitting variable factors into standard combinatoric property fractions simplifies coefficient conversions with geometric series denominators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula The general property linking indices to binomial coefficients is: r cdot binomnr = n cdot binomn-1r-1$r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1}$ ### Core Logic The given series can be structured using summation notation: S = sum_r=1^15 r^2 cdot binom15r$S = \sum_{r=1}^{15} r^2 \cdot \binom{15}{r}$ Apply the identity r cdot binom15r = 15 cdot binom14r-1$r \cdot \binom{15}{r} = 15 \cdot \binom{14}{r-1}$ to reduce one factor of r$r$: S = sum_r=1^15 r cdot left[ 15 cdot binom14r-1 right] = 15 sum_r=1^15 r cdot binom14r-1$S = \sum_{r=1}^{15} r \cdot \left[ 15 \cdot \binom{14}{r-1} \right] = 15 \sum_{r=1}^{15} r \cdot \binom{14}{r-1}$ ### Step 1: Splitting the linear term Rewrite the index variable r$r$ as (r - 1) + 1$(r - 1) + 1$ to align with the binomial lower index: S = 15 sum_r=1^15 big((r - 1) + 1big) cdot binom14r-1$S = 15 \sum_{r=1}^{15} \big((r - 1) + 1\big) \cdot \binom{14}{r-1}$ S = 15 sum_r=1^15 (r - 1) cdot binom14r-1 + 15 sum_r=1^15 binom14r-1$S = 15 \sum_{r=1}^{15} (r - 1) \cdot \binom{14}{r-1} + 15 \sum_{r=1}^{15} \binom{14}{r-1}$ Applying the property again to the first summation term: (r-1)binom14r-1 = 14binom13r-2$(r-1)\binom{14}{r-1} = 14\binom{13}{r-2}$: S = 15 cdot 14 sum_r=2^15 binom13r-2 + 15 sum_r=1^15 binom14r-1$S = 15 \cdot 14 \sum_{r=2}^{15} \binom{13}{r-2} + 15 \sum_{r=1}^{15} \binom{14}{r-1}$ ### Step 2: Evaluating the Sums and Prime Factorization Using the standard total sum of binomial coefficients sum_k=0^n binomnk = 2^n$\sum_{k=0}^{n} \binom{n}{k} = 2^n$: S = 15 cdot 14 cdot 2^13 + 15 cdot 2^14$S = 15 \cdot 14 \cdot 2^{13} + 15 \cdot 2^{14}$ Factor out 15 cdot 2^13$15 \cdot 2^{13}$ from the expression: S = 15 cdot 2^13 (14 + 2) = 15 cdot 2^13 (16) = 15 cdot 2^13 cdot 2^4$S = 15 \cdot 2^{13} (14 + 2) = 15 \cdot 2^{13} (16) = 15 \cdot 2^{13} \cdot 2^4$ S = 15 cdot 2^17 = (3^1 cdot 5^1) cdot 2^17$S = 15 \cdot 2^{17} = (3^1 \cdot 5^1) \cdot 2^{17}$ Matching this with the given format 2^m cdot 3^n cdot 5^k$2^m \cdot 3^n \cdot 5^k$, we identify: m = 17$m = 17$, n = 1$n = 1$, and k = 1$k = 1$. ### Step 3: Calculating the sum of exponents Evaluating the targeted summation: m + n + k = 17 + 1 + 1 = 19$m + n + k = 17 + 1 + 1 = 19$ ### Pattern Recognition For a series of the type sum r^2 binomnr$\sum r^2 \binom{n}{r}$, remember the standard identity shortcut: n(n-1)2^n-2 + n2^n-1$n(n-1)2^{n-2} + n2^{n-1}$. Plugging in n=15$n=15$ instantly outputs 15(14)2^13 + 15(2^14)$15(14)2^{13} + 15(2^{14})$, bypasses matching terms manually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem