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Question
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .
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Solution
Comparing the indices of x in xn and in Tr + 1, we obtain
r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
Hence, proved.
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