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For some n neq 10, let the coefficients of the 5^textth, 6^textth and 7^textth terms in the binomial expansion of (1 + x)^n+4 be in A.P. Then the largest coefficient in the expansion of (1 + x)^n+4 is :

Solution & Explanation

### Related Formula The coefficient of the r^textth term in the expansion (1+x)^m is written as binommr-1. For three terms in A.P., their values satisfy: 2 cdot T_2 = T_1 + T_3 ### Core Logic Let the total power exponent be m = n + 4. The coefficients of the 5^textth, 6^textth, and 7^textth terms are binomm4, binomm5, and binomm6 respectively. Since they form an arithmetic progression: 2 cdot binomm5 = binomm4 + binomm6 Rearrange using the recurrence addition identity properties: 4 cdot binomm5 = left[ binomm4 + binomm5 right] + left[ binomm5 + binomm6 right] 4 cdot binomm5 = binomm+15 + binomm+16 4 cdot binomm5 = binomm+26 ### Step 1: Solve the Combinatorial Fraction for m Expand the combinations using factorials: 4 cdot fracm!5!(m-4)! = frac(m+2)!6!(m-4)! Cancel out (m-4)! from both denominators: 4 cdot fracm!120 = frac(m+2)(m+1)m!720 4 = frac(m+2)(m+1)6 24 = m^2 + 3m + 2 implies m^2 + 3m - 22 = 0 Wait, let's re-verify the step using the direct ratio computation: 2 = fracbinomm4binomm5 + fracbinomm6binomm5 = frac5m-4 + fracm-56 2 = frac30 + (m-4)(m-5)6(m-4) 12(m-4) = 30 + m^2 - 9m + 20 12m - 48 = m^2 - 9m + 50 implies m^2 - 21m + 98 = 0 Factor the quadratic equation: (m-7)(m-14) = 0 implies m = 7 text or m = 14 ### Step 2: Connect back to the problem constraints Since m = n + 4: - If m = 14 implies n + 4 = 14 implies n = 10 (this is rejected because the problem states n neq 10). - If m = 7 implies n + 4 = 7 implies n = 3 (this value is accepted). Thus, the binomial expansion exponent is exactly m = 7. ### Step 3: Extract Maximum Binomial Coefficient For an odd exponent power m = 7, the maximum binomial coefficient corresponds to the middle terms: textMax Coefficient = binom73 = binom74 = frac7 cdot 6 cdot 53 cdot 2 cdot 1 = 35 ### Pattern Recognition For consecutive binomial coefficients binomnr-1, binomnr, binomnr+1 in A.P., the power parameters satisfy the standard identity (n-2r)^2 = n+2, which allows quick calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

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Binomial Theorem

More Binomial Theorem Previous-Year Questions — Page 4

Q68 2025 Properties of Coefficients and Rational Terms
Let the coefficients of three consecutive terms T_r, T_r+1 and T_r+2 in the binomial expansion of (a+b)^12 be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (sqrt[4]3+sqrt[3]4)^12 . Then p+q is equal to :
  • A. 283
  • B. 295
  • C. 287
  • D. 299

Solution

### Related Formula General term of binomial expansion (x + y)^n: T_k+1 = binomnk x^n-k y^k Condition for three terms A, B, C to be in G.P.: B^2 = A cdot C ### Core Logic Part 1: Coefficients of T_r, T_r+1, T_r+2 are binom12r-1, binom12r, binom12r+1. Since they are in G.P.: left[binom12rright]^2 = binom12r-1 cdot binom12r+1 fracbinom12rbinom12r-1 = fracbinom12r+1binom12r implies frac12-r+1r = frac12-rr+1 (13-r)(r+1) = r(12-r) implies 13r + 13 - r^2 - r = 12r - r^2 12r + 13 = 12r implies 13 = 0 quad (textNot possible) Thus, no real integer solution for r exists, so p = 0. ### Step 1: Calculate Rational Terms Sum q Part 2: Rational terms in expansion of (3^1/4 + 4^1/3)^12. General term: T_k+1 = binom12k (3^1/4)^12-k (4^1/3)^k = binom12k 3^frac12-k4 4^frack3 For the term to be rational, frac12-k4 and \frac{k}{3} must both be integers: - k must be a multiple of 3: k in \0, 3, 6, 9, 12\ - 12-k must be a multiple of 4, so k must be a multiple of 4: k in \0, 4, 8, 12\ The common values for k are k = 0 and k = 12. - At k = 0: T_1 = binom120 3^3 4^0 = 1 times 27 times 1 = 27 - At k = 12: T_13 = binom1212 3^0 4^4 = 1 times 1 times 256 = 256 Sum of rational terms q = 27 + 256 = 283. ### Step 2: Final Combination p + q = 0 + 283 = 283 ### Pattern Recognition To find common values for divisibility constraints, look for multiples of the least common multiple textlcm(3, 4) = 12 within the range [0, 12]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem
Q58 2025 Number of Integral Terms
The least value of n for which the number of integral terms in the Binomial expansion of left(sqrt[3]7 + sqrt[12]11right)^n is 183, is:
  • A. 2184
  • B. 2148
  • C. 2172
  • D. 2196

Solution

### Related Formula T_r+1 = binomnr a^n-r b^r ### Core Logic The general term in the expansion is: T_r+1 = binomnr (7)^fracn-r3 (11)^fracr12 For the term to be integral, both powers fracn-r3 and fracr12 must be integers. This dictates that r must be a multiple of 12 (r = 12k, where k is a non-negative integer). ### Step 1: Setup total count equation The values r can take are 0, 12, 24, dots, 12k. The total number of terms is given as 183. Since counting starts from k=0, the maximum value of k is: k_textmax = 183 - 1 = 182 ### Step 2: Calculate minimum n The maximum index value r required to achieve this count is: r_textmax = 12 times 182 = 2184 Hence, the least value of n must be 2184 to encompass all 183 integral terms. ### Pattern Recognition Number of terms formula when tracking steps of size L: textTotal Terms = lfloor fracnL rfloor + 1. Isolate n directly to compute bounds rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

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