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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 5

Q5 jee_main_2024_27_jan_morning Sum of Binomial Coefficients
If A denotes the sum of all the coefficients in the expansion of (1-3x+10x^2)^n and B denotes the sum of all the coefficients in the expansion of (1+x^2)^n, then :
  • A. A=B^3
  • B. 3A=B
  • C. B=A^3
  • D. A=3B

Solution

### Related Formula textSum of all coefficients in f(x)^n = f(1)^n ### Core Logic To find the sum of all coefficients in a polynomial expansion, substitute the variable x = 1. For A (Sum of coefficients of (1-3x+10x^2)^n): A = (1 - 3(1) + 10(1)^2)^n A = (1 - 3 + 10)^n = (8)^n For B (Sum of coefficients of (1+x^2)^n): B = (1 + (1)^2)^n B = (1 + 1)^n = (2)^n ### Step 1: Establishing the Relation We have A = 8^n and B = 2^n. Observe that 8 = 2^3. A = (2^3)^n = (2^n)^3 Substituting B = 2^n into the expression: A = B^3 ### Pattern Recognition Summing coefficients of *any* multi-nomial expansion is trivially simple: just plug in x=1 (or x=y=z=1 for multivariable expressions). It collapses the variables, leaving only the arithmetic sum of the coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem
Q29 jee_main_2024_29_jan_morning Sum of Binomial Coefficients
If frac^11C_12+frac^11C_23+dots+frac^11C_910=fracnm with gcd(n,m)=1, then n+m is equal to
Numerical Answer. Answer: 2041 to 2041

Solution

### Related Formula frac^nC_rr+1 = frac^n+1C_r+1n+1 sum_k=0^n ^nC_k = 2^n ### Core Logic The given series can be rewritten using summation notation: S = sum_r=1^9 frac^11C_rr+1 Applying the coefficient shifting identity frac^nC_rr+1 = frac^n+1C_r+1n+1: S = sum_r=1^9 frac^12C_r+112 S = frac112 sum_r=1^9 ^12C_r+1 ### Step 1: Expand and Complete the Series Expand the internal sum by shifting the index bounds: S = frac112 left( ^12C_2 + ^12C_3 + dots + ^12C_10 right) We know the complete sum of binomial coefficients for n=12 is 2^12. We just need to subtract the missing boundary terms: k = 0, 1, 11, 12. 2^12 = sum_k=0^12 ^12C_k The missing terms evaluate to: ^12C_0 = 1 ^12C_1 = 12 ^12C_11 = 12 ^12C_12 = 1 Sum of missing terms = 1 + 12 + 12 + 1 = 26. ### Step 2: Calculate Final Fraction Substitute this back into the series equation: S = frac112 left[ 2^12 - 26 right] S = frac112 [ 4096 - 26 ] S = frac407012 Simplify the fraction by dividing by 2 to achieve the coprime structure fracnm: S = frac20356 Thus, n = 2035 and m = 6, and they are coprime (gcd(2035, 6) = 1). Calculate n + m: n + m = 2035 + 6 = 2041 ### Pattern Recognition Whenever you see binomial coefficients divided by their sequential index (r+1), always use the absorption identity frac1n+1 binomn+1r+1 to bump the top index up by 1. Then fill the array to force the complete 2^n+1 sum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem
Q17 jee_main_2024_30_january_evening Binomial Coefficients
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
  • A. 4
  • B. 10
  • C. 8
  • D. 6

Solution

### 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
Q24 jee_main_2024_30_january_evening Properties of Binomial Coefficients
Let alpha = sum_k=0^nleft(frac(^nC_k)^2k+1right) and beta = sum_k=0^n-1left(frac^nC_k^nC_k+1k+2right) . If 5alpha = 6beta , then n equals
Numerical Answer. Answer: 10 to 10

Solution

### Related Formula frac^nC_rr+1 = frac^n+1C_r+1n+1 sum_k=0^n ^n+1C_k+1 cdot ^nC_n-k = ^2n+1C_n+1 ### Core Logic Simplify alpha by absorbing the denominator: alpha = sum_k=0^n frac^nC_k cdot ^nC_kk + 1 Multiply and divide by (n+1): alpha = frac1n+1 sum_k=0^n fracn+1k+1 ^nC_k cdot ^nC_n-k alpha = frac1n+1 sum_k=0^n ^n+1C_k+1 cdot ^nC_n-k This corresponds to choosing k+1 items from a set of n+1, and n-k items from a set of n. Total items chosen = (k+1) + (n-k) = n+1 from a total pool of (n+1) + n = 2n+1. alpha = frac1n+1 cdot ^2n+1C_n+1 ### Step 1: Evaluating Beta Now for beta: beta = sum_k=0^n-1 frac^nC_k cdot ^nC_k+1k + 2 Convert ^nC_k to ^nC_n-k and absorb the denominator into ^nC_k+1: beta = frac1n+1 sum_k=0^n-1 ^nC_n-k cdot fracn+1k+2 ^nC_k+1 beta = frac1n+1 sum_k=0^n-1 ^nC_n-k cdot ^n+1C_k+2 Here, total items chosen is (n-k) + (k+2) = n+2 from a total pool of n + (n+1) = 2n+1. beta = frac1n+1 cdot ^2n+1C_n+2 ### Step 2: Applying the Given Ratio Given 5alpha = 6beta Rightarrow fracbetaalpha = frac56. fracbetaalpha = fracfrac1n+1 ^2n+1C_n+2frac1n+1 ^2n+1C_n+1 = frac^2n+1C_n+2^2n+1C_n+1 Using the ratio property frac^nC_r^nC_r-1 = fracn-r+1r: Here, n to 2n+1 and r to n+2. fracbetaalpha = frac2n+1 - (n+2) + 1n+2 = frac2n+1 - n - 2 + 1n+2 = fracnn+2 ### Step 3: Solving for n Equate and solve: fracnn+2 = frac56 6n = 5n + 10 Rightarrow n = 10 ### Pattern Recognition Any fractional binomial coefficient like C_k / (k+x) triggers an absorption identity to step up the index. Then, multiplying terms turns into a simple combinatorial Vandermonde convolution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem
Q26 jee_main_2024_30_jan_morning General Term
Number of integral terms in the expansion of left\7^left(frac12right) + 11^left(frac16right)right\^824 is equal to
Numerical Answer. Answer: 138 to 138

Solution

### Related Formula T_r+1 = binomnr a^n-r b^r ### Core Logic General term in the expansion of left(7^1/2 + 11^1/6right)^824 is: t_r+1 = binom824r (7)^frac824-r2 (11)^r/6 For the term to be integral, both powers must be integers. This means: 1) frac824 - r2 must be an integer, which means r must be even. 2) fracr6 must be an integer, which means r must be a multiple of 6. Since any multiple of 6 is already even, the condition reduces to: r must be a multiple of 6. ### Step 1: Finding valid values of r The possible values for r are 0, 1, 2, dots, 824. Valid r = 0, 6, 12, dots, 822. This forms an arithmetic progression with first term a=0, common difference d=6, and last term L=822. L = a + (n-1)d 822 = 0 + (n-1)6 n - 1 = frac8226 = 137 n = 138 Thus, there are 138 integral terms. ### Pattern Recognition Finding rational/integral terms in a binomial expansion strictly requires finding the LCM of the fractional power denominators, then counting multiples up to n. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

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