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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

More Binomial Theorem Previous-Year Questions

Q65 2025 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 2025 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 2025 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 2025 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 2025 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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