If the coefficient of x in the expansion of (ax^2 + bx + c)(1 - 2x)^26 is -56 and the coefficients of x^2 and x^3 are both zero, then a + b + c is equal to

Solution & Explanation

### Related Formula (1 - 2x)^26 = sum_r=0^26 binom26r (-2x)^r The coefficient extraction utilizes the distributive property over the polynomial (ax^2+bx+c). ### Core Logic Expansion of expression: (ax^2 + bx + c) sum_r=0^26 binom26r (-2x)^r For the term in 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 binom260(-2)^0 + b cdot binom261(-2)^1 + c cdot binom262(-2)^2 = 0 a - 52b + 1300c = 0 dots(1) ### Step 1: Formulate the system of linear equations For the term in 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 binom261(-2)^1 + b cdot binom262(-2)^2 + c cdot binom263(-2)^3 = 0 -52a + 1300b - 20800c = 0 dots(2) For the term in x^1 (set to -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 - 52c = -56 dots(3) ### Step 2: Solve the Linear System From (3), b = 52c - 56. Substitute into (2) divided by -52 to simplify: a - 25b + 400c = 0. Substitute into (1): a - 52b + 1300c = 0. Subtracting: 27b - 900c = 0 Rightarrow b = frac1003c. Wait, recalculating directly: From (2): -52a + 1300b - 20800c = 0 Rightarrow a - 25b + 400c = 0. From (1): a - 52b + 1300c = 0. (a - 25b + 400c) - (a - 52b + 1300c) = 0 Rightarrow 27b - 900c = 0 Rightarrow 3b = 100c. Substitute into (3): 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. Divide by -52: a - 25b + 400c = 0. Correct. (a - 52b + 1300c) - (a - 25b + 400c) = -27b + 900c = 0 Rightarrow 27b = 900c Rightarrow b = frac1003c. This contradicts integer expectations. Let's substitute b from (3) directly. b = 52c - 56. 27(52c - 56) = 900c 1404c - 1512 = 900c Rightarrow 504c = 1512 Rightarrow c = 3. Then b = 52(3) - 56 = 156 - 56 = 100. From (1): a - 52(100) + 1300(3) = 0 Rightarrow a - 5200 + 3900 = 0 Rightarrow a = 1300. ### Step 3: Compute final value We have a = 1300, b = 100, c = 3. 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) upward sequentially unwinds the system cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

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Binomial Theorem

More Binomial Theorem Previous-Year Questions — Page 2

Q66 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)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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