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Question
Find the coefficient of:
(vi) x in the expansion of \[\left( 1 - 2 x^3 + 3 x^5 \right) \left( 1 + \frac{1}{x} \right)^8\]
Solution
Suppose x occurs at the (r + 1)th term in the given expression.
Then, we have:
\[(1 - 2 x^3 + 3 x^5 ) \left( 1 + \frac{1}{x} \right)^8 \]
\[ = \left( 1 - 2 x^3 + 3 x^5 \right)\left( ^{8}{}{C}_0 + ^{8}{}{C}_1 \left( \frac{1}{x} \right) + ^{8}{}{C}_2 \left( \frac{1}{x} \right)^2 + ^{8}{}{C}_3 \left( \frac{1}{x} \right)^3 +^{8}{}{C}_4 \left( \frac{1}{x} \right)^4 + ^{8}{}{C}_5 \left( \frac{1}{x} \right)^5 + ^{8}{}{C}_6 \left( \frac{1}{x} \right)^6 +^{8}{}{C}_7 \left( \frac{1}{x} \right)^7 +^{8}{}{C}_8 \left( \frac{1}{x} \right)^8 \right)\]
\[ \text{ x occurs in the above expresssion at } - 2 x^3 . ^{8}{}{C}_2 \left( \frac{1}{x^2} \right) + 3 x^5 . ^{8}{}{C}_4 \left( \frac{1}{x} \right)^4 . \]
\[ \therefore \text{ Coefficient of x } = - 2\left( \frac{8!}{2! 6!} \right) + 3\left( \frac{8!}{4! 4!} \right) = - 56 + 210 = 154\]
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