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Question : 63 of 160
Marks:
+1,
-0
Solution:
| (1−x)(1−x2)...(1−x2n) |
| {(1−x)(1−x2)...(1−xn)}2 |
=| (x−1)(x2−1)...(x2n−1) |
| {(x−1)(x2−1)...(xn−1)}2 |
=| (x−1)(x2−1)...(x2n−1) |
| {..}2.(x−1)2n |
= = == =2nCn [∵nan−1]
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