Suppose 2 - p, p, 2 - alpha, alpha are the coefficient of four consecutive terms in the expansion of (1 + x)^n . Then the value of p^2 - alpha^2 + 6alpha + 2p equals

Solution & Explanation

### Related Formula textCoefficient of (r+1)^textth text term in (1+x)^n text is ^nC_r ^nC_r + ^nC_r+1 = ^n+1C_r+1 ### Core Logic Let the four consecutive binomial coefficients be ^nC_r, ^nC_r+1, ^nC_r+2, ^nC_r+3. Given these correspond to 2 - p, p, 2 - alpha, alpha respectively. From the properties of combinations: ^nC_r + ^nC_r+1 = (2 - p) + p = 2 Rightarrow ^n+1C_r+1 = 2 quad dots(1) Similarly: ^nC_r+2 + ^nC_r+3 = (2 - alpha) + alpha = 2 Rightarrow ^n+1C_r+3 = 2 quad dots(2) ### Step 1: Identifying the Inconsistency From equations (1) and (2): ^n+1C_r+1 = ^n+1C_r+3 = 2 By combination properties, if ^nC_x = ^nC_y and x neq y, then x + y = n. So, (r+1) + (r+3) = n+1 Rightarrow 2r + 4 = n+1 Rightarrow n = 2r + 3. Substitute n back into the equality: ^2r+4C_r+1 = 2 The binomial coefficient must be ge 2. However, for any valid integer r ge 0, ^2r+4C_r+1 grows very rapidly. Let's test small values: If r = 0, ^4C_1 = 4 neq 2. If r = 1, ^6C_2 = 15 neq 2. Hence, no integer values satisfy this condition, making the given data inherently inconsistent. ### Step 2: Conclusion Due to inconsistent data resulting in a mathematically impossible scenario, this question was treated as a Bonus/Dropped question. ### Pattern Recognition Adding consecutive binomial coefficients yields the sum from Pascal's triangle identity ^nC_r + ^nC_r+1 = ^n+1C_r+1. If identical sums yield small integer invariants like 2, they typically break bounding limits for ^nC_k combinations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

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Q jee_main_2025_08_april_evening Integral Terms in Binomial Expansion
The number of integral terms in the expansion of left(5^frac12 + 7^frac18right)^1016 is
  • A. 127
  • B. 130
  • C. 129
  • D. 128

Solution

### Related Formula T_r = binomnr a^n-r b^r ### Core Logic Formulate general progression steps. Isolate divisor common multiples to find non fraction power configurations tracking perfectly over boundaries. ### Step 1: State the General Binomial Term T_r = binom1016r 5^frac1016-r2 7^fracr8 ### Step 2: Apply Rational Power Constraints For expressions to map purely to integer categories, indexing tracker index paths r must explicitly scale as multiples of 8 over the domain range: r in \0, 8, 16, 24, dots, 1016\ ### Step 3: Enumerate Arithmetic Progression Size Using standard progression length mapping tools: 1016 = 0 + (n - 1) cdot 8 n - 1 = frac10168 = 127 implies n = 128 ### Pattern Recognition Finding pure integer steps matches finding values that fit LCM tracking parameters for base radical roots across whole block lengths. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem
Q74 jee_main_2025_08_april_evening Last Two Digits of a Number
The product of the last two digits of (1919)^1919 is
Numerical Answer. Answer: 63 to 63

Solution

### Related Formula (10k - 1)^n = dots + binomnn-1(10k)(-1)^n-1 + (-1)^n ### Core Logic Break the base parameter into a multiple of 10 form (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 = binom19190(1920)^1919 - dots + binom19191918(1920)^1 - binom19191919 ### 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 ### Step 3: Deduce Final Trailing Quotient Product 1919 times 1920 - 1 = 3684480 - 1 = 3684479 Trailing structural indicators: 79. Product calculation response: 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
Q67 jee_main_2025_29_jan_evening Remainder Problems
The remainder, when 7^103 is divided by 23, is equal to:
  • A. 14
  • B. 9
  • C. 17
  • D. 6

Solution

### Related Formula Modular arithmetic congruence rules: A equiv B pmodm implies A^n equiv B^n pmodm ### Core Logic We need to compute 7^103 pmod23. Let's break the exponent down into manageable powers: 7^103 = 7 cdot (7^2)^51 = 7 cdot (49)^51 Since 49 equiv 3 pmod23: 7^103 equiv 7 cdot 3^51 pmod23 ### Step 1: Simplify Higher Power Components Break down 3^51 using 3^3 = 27 equiv 4 pmod23: 3^51 = (3^3)^17 = 27^17 equiv 4^17 pmod23 So the expression becomes: 7 cdot 4^17 = 7 cdot 4 cdot (4^2)^8 = 28 cdot 16^8 Since 28 equiv 5 pmod23 and 16 equiv -7 pmod23: equiv 5 cdot (-7)^8 = 5 cdot 7^8 pmod23 ### Step 2: Final Remainder Evaluation Now compute 7^8 using 7^2 = 49 equiv 3 pmod23: 7^8 = (7^2)^4 equiv 3^4 = 81 pmod23 Since 81 equiv 12 pmod23: textExpression equiv 5 cdot 12 = 60 pmod23 Dividing 60 by 23 (23 times 2 = 46) yields a remainder of: 60 - 46 = 14 ### Pattern Recognition Using negative remainders (like 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
Q54 jee_main_2025_03_april_morning Rational Terms in Expansion
The sum of all rational terms in the expansion of (2 + sqrt3)^8 is[cite: 540, 542]:
  • A. 16923
  • B. 3763
  • C. 33845
  • D. 18817

Solution

### Related Formula General term in the binomial expansion of (a+b)^n: T_r+1 = ^nC_r cdot a^n-r cdot b^r ### Core Logic For a term to be rational in (2+sqrt3)^8, the power of sqrt3 must be an even integer[cite: 1218]. Thus, r can only take even values from 0 to 8 [cite: 1219, 1220]: 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 [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 [cite: 1219, 1220] Compute the individual numeric values [cite: 1221]: - r=0: 1 cdot 256 = 256 - r=2: 28 cdot 64 cdot 3 = 5376 - r=4: 70 cdot 16 cdot 9 = 10080 - r=28 cdot 4 cdot 27 = 3024 - r=8: 1 cdot 1 cdot 81 = 81 textTotal 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem
Q59 jee_main_2025_03_april_morning Binomial Coefficients Series
If sum_r=1^9left(fracr + 32^rright)cdot^9mathrmC_r = alpha left(frac32right)^9 - beta [cite: 596], where alpha, beta in mathbbN [cite: 596], then (alpha + beta)^2 is equal to[cite: 608]:
  • A. 27
  • B. 9
  • C. 81
  • D. 18

Solution

### Related Formula 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 ### 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 [cite: 1316] Simplify the first sub-sum using the index relation 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 [cite: 1316] Simplify the second sub-sum by including missing index r=0 [cite: 1316]: 3left[sum_r=0^9 ^9C_rleft(frac12right)^r - 1right] = 3left[left(1 + frac12right)^9 - 1right] = 3left(frac32right)^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 [cite: 1316] Convert the fractional leading term [cite: 1316]: frac92left(frac32right)^8 = 3 cdot frac32left(frac32right)^8 = 3left(frac32right)^9 [cite: 1316] textTotal Sum = 3left(frac32right)^9 + 3left(frac32right)^9 - 3 = 6left(frac32right)^9 - 3 [cite: 1316] Matching coefficients gives [cite: 1317]: alpha = 6, quad beta = 3 [cite: 1317] Evaluate the required squared value [cite: 1317]: (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

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