JEE Maths›Binomial Theorem›Visual Solution
Visual SolutionPYQ 2024 · Jan 29 Shift 1Tricky
Question
If 11C12+11C23+…+11C910=nm\frac{^{11}C_1}{2} + \frac{^{11}C_2}{3} + \ldots + \frac{^{11}C_9}{10} = \frac{n}{m} with gcd⁡(n,m)=1\gcd(n, m) = 1, then n+mn + m is equal to _____.
Solution Path
Absorption identity: nCrr+1=n+1Cr+1n+1\frac{{}^{n}C_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}. Sum =407012=20356= \frac{4070}{12} = \frac{2035}{6}. n+m=2041n + m = 2041.
01Question Setup
1/4
Find n+mn + m where 11C12+11C23+⋯+11C910=nm\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \cdots + \frac{{}^{11}C_9}{10} = \frac{n}{m} with gcd⁡(n,m)=1\gcd(n,m) = 1.
n+m=  ?n + m = \;?
02Absorption Identity
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Use nCrr+1=n+1Cr+1n+1\frac{{}^{n}C_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1} to convert the sum into 112∑k=21012Ck\frac{1}{12}\sum_{k=2}^{10} {}^{12}C_k.
112∑k=21012Ck\frac{1}{12}\sum_{k=2}^{10} {}^{12}C_k
03Evaluate SumKEY INSIGHT
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∑k=21012Ck=212−1−12−12−1=4070\sum_{k=2}^{10} {}^{12}C_k = 2^{12} - 1 - 12 - 12 - 1 = 4070. Sum =407012=20356= \frac{4070}{12} = \frac{2035}{6}.
20356\frac{2035}{6}
04Final Answer
4/4
n=2035n = 2035, m=6m = 6, so n+m=2041n + m = 2041.
2041\boxed{2041}
Concepts from this question1 concepts unlocked
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Sum of Coefficients via Substitution

EASY

Sum of all coefficients of f(x) = f(1). Sum of even-indexed = (f(1)+f(-1))/2

∑coefficients=f(1),∑even=f(1)+f(−1)2\sum \text{coefficients} = f(1), \quad \sum \text{even} = \frac{f(1)+f(-1)}{2}

Instant 30-second solve for coefficient sum questions. No expansion needed.

Coefficient problemsSum/difference of coefficientsSubstitution trick
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