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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

Reference Study Guides

Binomial Theorem

More Binomial Theorem Previous-Year Questions — Page 6

Q24 jee_main_2024_31_jan_evening Series Expansion
Let the coefficient of x^r in the expansion of (x+3)^n-1 + (x+3)^n-2(x+2) + (x+3)^n-3(x+2)^2 + dots + (x+2)^n-1 be alpha_r. If sum_r=0^n alpha_r = beta^n - gamma^n, where beta, gamma in mathbbN, then the value of beta^2 + gamma^2 equals
Numerical Answer. Answer: 25 to 25

Solution

### Related Formula textSum 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 ### Core Logic Let the expanded polynomial be P(x). The sum of its coefficients is sum alpha_r = P(1). Substitute x = 1 into the given expression: 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 and common ratio r = 3/4. There are 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 Comparing this with beta^n - gamma^n, we get: beta = 4, quad gamma = 3 Calculate beta^2 + gamma^2: beta^2 + gamma^2 = 4^2 + 3^2 = 16 + 9 = 25 ### Pattern Recognition Substituting x=1 immediately bypasses expanding individual 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
Q2 jee_main_2024_31_jan_morning Sum of Coefficients and Limits
Let a be the sum of all coefficients in the expansion of (1 - 2x + 2x^2)^2023 (3 - 4x^2 + 2x^3)^2024 and b = lim_x to 0 left( fracint_0^x fraclog(1 + t)t^2024 + 1 dtx^2 right). If the equations cx^2 + dx + e = 0 and 2bx^2 + ax + 4 = 0 have a common root, where c, d, e in mathbbR, then d : c : e equals
  • A. 2:1:4
  • B. 4:1:4
  • C. 1:2:4
  • D. 1:1:4

Solution

### Core Logic To find the sum of all coefficients in a polynomial expansion, substitute x = 1. a = (1 - 2(1) + 2(1)^2)^2023 (3 - 4(1)^2 + 2(1)^3)^2024 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 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 = 1 times frac12 = frac12 ### Step 2: Analyze Common Roots The given second equation is 2bx^2 + ax + 4 = 0. Substitute a = 1 and b = frac12: 2left(frac12right)x^2 + 1(x) + 4 = 0 implies x^2 + x + 4 = 0 The discriminant of x^2 + x + 4 = 0 is D = 1 - 16 < 0. Roots are non-real complex conjugates. ### Step 3: Final Ratio Since c, d, e in mathbbR 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 This implies 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), 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
Q25 jee_main_2024_31_jan_morning Coefficients in Expansion
In the expansion of (1 + x)(1 - x^2)left(1 + frac3x + frac3x^2 + frac1x^3right)^5, x neq 0, the sum of the coefficient of x^3 and x^-13 is equal to
Numerical Answer. Answer: 118 to 118

Solution

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

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