If 1^2 cdot left( ^15 mathrmC_1 right) + 2^2 cdot left( ^15 mathrmC_2 right) + 3^2 cdot left( ^15 mathrmC_3 right) + dots + 15^2 cdot left( ^15 mathrmC_15 right) = 2^mathrmm cdot 3^mathrmn cdot 5^mathrmk, where m, n, k ∈ N, then m + n + k is equal to :-

Solution & Explanation

### Related Formula The general property linking indices to binomial coefficients is: r cdot binomnr = n cdot binomn-1r-1 ### Core Logic The given series can be structured using summation notation: S = sum_r=1^15 r^2 cdot binom15r Apply the identity r cdot binom15r = 15 cdot binom14r-1 to reduce one factor of r: S = sum_r=1^15 r cdot left[ 15 cdot binom14r-1 right] = 15 sum_r=1^15 r cdot binom14r-1 ### Step 1: Splitting the linear term Rewrite the index variable r as (r - 1) + 1 to align with the binomial lower index: S = 15 sum_r=1^15 big((r - 1) + 1big) cdot binom14r-1 S = 15 sum_r=1^15 (r - 1) cdot binom14r-1 + 15 sum_r=1^15 binom14r-1 Applying the property again to the first summation term: (r-1)binom14r-1 = 14binom13r-2: S = 15 cdot 14 sum_r=2^15 binom13r-2 + 15 sum_r=1^15 binom14r-1 ### Step 2: Evaluating the Sums and Prime Factorization Using the standard total sum of binomial coefficients sum_k=0^n binomnk = 2^n: S = 15 cdot 14 cdot 2^13 + 15 cdot 2^14 Factor out 15 cdot 2^13 from the expression: S = 15 cdot 2^13 (14 + 2) = 15 cdot 2^13 (16) = 15 cdot 2^13 cdot 2^4 S = 15 cdot 2^17 = (3^1 cdot 5^1) cdot 2^17 Matching this with the given format 2^m cdot 3^n cdot 5^k, we identify: m = 17, n = 1, and k = 1. ### Step 3: Calculating the sum of exponents Evaluating the targeted summation: m + n + k = 17 + 1 + 1 = 19 ### Pattern Recognition For a series of the type sum r^2 binomnr, remember the standard identity shortcut: n(n-1)2^n-2 + n2^n-1. Plugging in n=15 instantly outputs 15(14)2^13 + 15(2^14), bypasses matching terms manually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

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Binomial Theorem

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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
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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