Coefficient Comparison
Suppose the coefficients of three consecutive terms in the expansion of (1 + x)^n are in the ratio 1 : 7 : 35; from this information the value of n can be deduced to be which number?
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About This Question
- Subject
- mathematics
- Chapter
- binomial theorem and its simple applications
- Topic
- coefficient comparison
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
23
Three consecutive coefficients in (1 + x)^n are C(n, r-1), C(n, r), C(n, r+1), and equating their successive ratios to the given numbers produces a solvable system, a standard JEE Advanced setup. The ratio of the first two coefficients gives C(n, r)/C(n, r-1) = (n - r + 1)/r = 7/1 = 7, so n - r + 1 = 7r, hence n = 8r - 1. The ratio of the next two gives C(n, r+1)/C(n, r) = (n - r)/(r + 1) = 35/7 = 5, so n - r = 5(r + 1), hence n = 6r + 5. Equating 8r - 1 = 6r + 5 yields 2r = 6, so r = 3, and then n = 8(3) - 1 = 23. Option 21 results from misreading the second ratio as 35/1. Option 35 confuses a ratio entry with n itself. Option 17 comes from an index slip. The two consecutive-ratio equations uniquely fix both r and n. Plausibility check: with n = 23, r = 3 the coefficients C(23,2):C(23,3):C(23,4) = 253:1771:8855 reduce to 1:7:35, exactly the stated ratio.
This hard difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of coefficient comparison. It appeared in the 2025 exam.
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