Question
The fourth term in the expansion of (1−2x)32 will be
Answer: Option B
:
B
Expansion of (1−2x)32
= 1+32(−2x)+32.12.12(−2x)2+32.12(−12)16(−2x)3+.........
Hence 4th term is x32
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:
B
Expansion of (1−2x)32
= 1+32(−2x)+32.12.12(−2x)2+32.12(−12)16(−2x)3+.........
Hence 4th term is x32
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