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If the Coefficient of x^30 in the expansion of left(1+frac1xright)^6(1+x^2)^7(1-x^3)^8, xne0 is alpha, then |alpha| equals

Numerical Answer Type:
Enter a numerical value Answer: 678 to 678 +4 marks

Solution & Explanation

### Related Formula General term in a binomial expansion (1+t)^n is given by: T_r+1 = binomnr t^r ### Core Logic Let's simplify the algebraic structure of the product expression first: left(1+frac1xright)^6(1+x^2)^7(1-x^3)^8 = frac(x+1)^6 (1+x^2)^7 (1-x^3)^8x^6 Finding the coefficient of x^30 in this full product is equivalent to finding the coefficient of x^36 in the numerator expansion: textTarget = textCoefficient of x^36 text in (1+x)^6 (1+x^2)^7 (1-x^3)^8 ### Step 1: Setting up General Term Constraints The product of the three general terms is: binom6r_1 x^r_1 cdot binom7r_2 (x^2)^r_2 cdot binom8r_3 (-x^3)^r_3 = binom6r_1binom7r_2binom8r_3 (-1)^r_3 x^r_1 + 2r_2 + 3r_3 We require the total exponent to equal 36: r_1 + 2r_2 + 3r_3 = 36 with boundaries 0 le r_1 le 6, 0 le r_2 le 7, 0 le r_3 le 8. ### Step 2: Case Analysis by r3 Let's evaluate non-vanishing integer combinations case-by-case: - **Case I: r_3 = 8** r_1 + 2r_2 = 36 - 24 = 12 - r_2 = 6, r_1 = 0 implies binom60binom76binom88(-1)^8 = 1 times 7 times 1 = 7 - r_2 = 5, r_1 = 2 implies binom62binom75binom88(-1)^8 = 15 times 21 times 1 = 315 - r_2 = 4, r_1 = 4 implies binom64binom74binom88(-1)^8 = 15 times 35 times 1 = 525 - r_2 = 3, r_1 = 6 implies binom66binom73binom88(-1)^8 = 1 times 35 times 1 = 35 - **Case II: r_3 = 7** r_1 + 2r_2 = 36 - 21 = 15 - r_2 = 7, r_1 = 1 implies binom61binom77binom87(-1)^7 = 6 times 1 times 8 times (-1) = -48 - r_2 = 6, r_1 = 3 implies binom63binom76binom87(-1)^7 = 20 times 7 times 8 times (-1) = -1120 - r_2 = 5, r_1 = 5 implies binom65binom75binom87(-1)^7 = 6 times 21 times 8 times (-1) = -1008 - **Case III: r_3 = 6** r_1 + 2r_2 = 36 - 18 = 18 - r_2 = 7, r_1 = 4 implies binom64binom77binom86(-1)^6 = 15 times 1 times 28 = 420 - r_2 = 6, r_1 = 6 implies binom66binom76binom86(-1)^6 = 1 times 7 times 28 = 196 ### Step 3: Summation for Alpha Summing all calculated values: alpha = (7 + 315 + 525 + 35) + (-48 - 1120 - 1008) + (420 + 196) alpha = 882 - 2176 + 616 = -678 Thus, the absolute value is: |alpha| = 678 ### Pattern Recognition Sees: Multi-product polynomial coefficient extraction problem. Trap: Remember that the negative sign inside (1-x^3)^8 alters the polarity of terms based on whether r_3 is odd or even. Always track the (-1)^r_3 factor carefully. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

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Binomial Theorem

More Binomial Theorem Previous-Year Questions

Q65 jee_main_2025_02_april_evening Properties of Binomial Coefficients
If sum_mathrmr = 0^10left(frac10^mathrmr + 1 - 110^mathrmrright) cdot binom11r + 1 = fracalpha^11 - 11^1110^10, then alpha is equal to:
  • A. 15
  • B. 11
  • C. 24
  • D. 20

Solution

### Related Formula textSum of binomial coefficients: sum_k=0^n binomnk x^k = (1 + x)^n textShifted binomial sum: sum_k=1^n binomnk 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. ### 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 Substitute this back into the sum: S = sum_r=0^10 left[ 10 - 10 left( frac110 right)^r+1 right] binom11r+1 S = 10 sum_r=0^10 binom11r+1 - 10 sum_r=0^10 binom11r+1 left( frac110 right)^r+1 ### Step 2: Evaluate both parts of the sum For the first part, let s = r+1: sum_r=0^10 binom11r+1 = sum_s=1^11 binom11s = 2^11 - 1 For the second part, using 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 ### 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 cdot 2^11 - 10 - 10 cdot frac11^1110^11 + 10 = 10 cdot 2^11 - frac11^1110^10 Express the first term with a denominator of 10^10: 10 cdot 2^11 = frac10^11 cdot 2^1110^10 = frac20^1110^10 Thus, the total sum is: S = frac20^11 - 11^1110^10 Comparing this with fracalpha^11 - 11^1110^10, we find: alpha = 20 ### Pattern Recognition Binomial base scaling: Whenever you see a sum of the form sum a^r binomnr, it is simply the expanded form of a shifted binomial expansion of (1 + a)^n. Factoring out scaling constants yields standard analytical forms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem
Q jee_main_2025_02_april_morning Term Independent of x
The term independent of x in the expansion of left(frac (x + 1)(x ^ frac 23 + 1 - x ^ frac 13) - frac (x - 1)(x - x ^ frac 12)right) ^ 1 0, x > 1 is:
  • A. 210
  • B. 150
  • C. 240
  • D. 120

Solution

### 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(sqrtxright)^2 - 1^2 = (sqrtx-1)(sqrtx+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 Second fraction: fracx-1x-x^1/2 = frac(sqrtx-1)(sqrtx+1)sqrtx(sqrtx-1) = fracsqrtx+1sqrtx = 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 ### Step 2: Apply Binomial Theorem The expression simplifies to left(x^1/3 - x^-1/2right)^10. Write the general term 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 ### Step 3: Solve for Independent Term Set the net power of x to zero: frac10-r3 - fracr2 = 0 implies 20 - 2r - 3r = 0 implies 5r = 20 implies r = 4 Substitute 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 ### Pattern Recognition The initial fractions look complex but contain simple hidden identities (a^3+b^3 and 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
Q72 jee_main_2025_03_april_evening Trinomial Expansion
Let (1 + x + x^2)^10 = a_0 + a_1x + a_2x^2 + dots + a_20x^20. If (a_1 + a_3 + a_5 + dots + a_19) - 11a_2 = 121k, then k is equal to
Numerical Answer. Answer: 239 to 239

Solution

### Related Formula For trinomial expansion: (1+x+x^2)^n = sum_r=0^2n a_r x^r Symmetric coefficients can be extracted by substituting x=1 and x=-1. ### Core Logic Let's perform the substitution: - For x = 1: 3^10 = a_0 + a_1 + a_2 + dots + a_20 quad text--- (1) - For x = -1: 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) a_1 + a_3 + a_5 + dots + a_19 = frac3^10 - 12 Since 3^5 = 243 implies 3^10 = 59049: a_1 + a_3 + a_5 + dots + a_19 = frac59049 - 12 = 29524 ### Step 2: Finding a_2 and k Using expansion formula to find a_2 (coefficient of x^2): (1 + x + x^2)^10 = sum frac10!p! q! r! x^q + 2r quad (p+q+r = 10) For q + 2r = 2: - Case 1: r=1, q=0 implies p=9. Coefficient = frac10!9! 0! 1! = 10 - Case 2: r=0, q=2 implies p=8. Coefficient = frac10!8! 2! 0! = 45 a_2 = 10 + 45 = 55 Now substitute into the target relation: 29524 - 11(55) = 29524 - 605 = 28919 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) rather than standard multinomial permutations to avoid calculation mistakes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem
Q54 jee_main_2025_07_april_morning Remainder Problems
The remainder when left((64)^(64)right)^(64) is divided by 7 is equal to
  • A. 4
  • B. 1
  • C. 3
  • D. 6

Solution

### Related Formula Binomial expansion for checking remainders: (1 + kx)^n = 1 + n(kx) + fracn(n-1)2(kx)^2 + dots equiv 1 pmod x ### Core Logic Let the target expression be N = left((64)^64right)^64. Using power rules, N = 64^64 times 64 = 64^64^2. Let the large exponent exponent be n = 64^2. We need to find the remainder of 64^n when divided by 7. ### Step 1: Express Base in Terms of Modulus Observe that 64 = 63 + 1 = 7 times 9 + 1. Substituting this into the expression: 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 + 63 cdot lambda = 1 + 7(9lambda) Since 7(9lambda) is perfectly divisible by 7, the remaining term is 1. ### Pattern Recognition Whenever the base can be written as km + 1, where m is the divisor, the value of (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
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

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