General Term
The coefficient of x¹₂ in the expansion of (1 + x)⁶(1 + x²)⁷ is
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About This Question
- Subject
- mathematics
- Chapter
- binomial theorem and its simple applications
- Topic
- general term
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
882
- Series Expansion:
- (1+x)6=∑r=06(r6)xr
- (1+x2)7=∑s=07(s7)x2s
- Compound Term: We need xk where k=r+2s=12.
- Valid Pairs (r,s):
- s=6⟹r=0:(06)(67)=1×7=7
- s=5⟹r=2:(26)(57)=15×21=315
- s=4⟹r=4:(46)(47)=15×35=525
- s=3⟹r=6:(66)(37)=1×35=35
- Summation: Total =7+315+525+35=882.
This hard difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of general term. It appeared in the 2025 exam.
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