Given:
(1 – 3x + 7x2) (1 – x)16
If x occurs at the (r + 1)th term in the given expression.
Then, we have:
(1 – 3x + 7x2) (1 – x)16 = (1 – 3x + 7x2) (16C0 + 16C1 (-x) + 16C2 (-x)2 + 16C3 (-x)3 + 16C4 (-x)4 + 16C5 (-x)5 + 16C6 (-x)6 + 16C7 (-x)7 + 16C8 (-x)8 + 16C9 (-x)9 + 16C10 (-x)10 + 16C11 (-x)11 + 16C12 (-x)12 + 16C13 (-x)13 + 16C14 (-x)14 + 16C15 (-x)15 + 16C16 (-x)16)
So, ‘x’ occurs in the above expression at 16C1 (-x) – 3x16C0
Coefficient of x = \(\dfrac{ -16!}{(1! 15!)} – 3(\dfrac{16!}{(0! 16!)})\)
= -16 – 3
= -19
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