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Question

The 4th term from the end in the expansion of (x32−2x2)7 is

A
35x
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B
70x2
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C
35x2
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D
70x
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Solution

The correct option is D 70x
For the above question
Tr+1=7Crx215r22r7
For the fourth term, from the end r=4
T5+1=7C4x12
=(35)(2)x
=70x

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A+B+C=N (N Fixed)
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