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In the expansion of (x2 – 1/x2)16, the value of constant term is_________________.
Answers (1)
Here, Tr+1 = 16Cr(x2)16-r(-1/x2)r
= 16Crx32-4r(-1)r
Now, for constant term,
32 – 4r = 0
Thus, r = 8
Thus, T8+1 = 16C8
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