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Question
Using binomial theorem, write down the expansions :
(viii) \[\left( 1 + 2x - 3 x^2 \right)^5\]
Solution
(viii) \[(1 + 2x - 3 x^2 )^5 \]
\[\text{ Consider 1 - 2x and } 3 x^2 \text{ as two separate entities and apply the binomial theorem } . \]
\[\text{ Now } , \]
\[^{5}{}{C}_0 (1 + 2x )^5 (3x )^0 -^{5}{}{C}_1 (1 + 2x )^4 (3 x^2 )^1 + ^{5}{}{C}_2 (1 + 2x )^3 (3 x^2 )^2 -^{5}{}{C}_3 (1 + 2x )^2 (3 x^2 )^3 + ^{5}{}{C}_4 (1 + 2x )^1 (3 x^2 )^4 -^{5}{}{C}_5 (1 + 2x )^0 (3 x^2 )^5 \]
\[ = (1 + 2x )^5 - 5(1 + 2x )^4 \times 3 x^2 + 10 \times (1 + 2x )^3 \times 9 x^4 - 10 \times (1 + 2x )^2 \times 27 x^6 + 5(1 + 2x) \times 81 x^8 - 243 x^{10} \]
\[ = ^{5}{}{C}_0 \times (2x )^0 + ^{5}{}{C}_1 \times (2x )^1 + ^{5}{}{C}_2 \times (2x )^2 + ^{5}{}{C}_3 \times (2x )^3 + ^{5}{}{C}_4 \times (2x )^4 +^{5}{}{C}_5 \times (2x )^5 - \]
\[ 15 x^2 [ ^{4}{}{C}_0 (2x )^0 + ^{4}{}{C}_1 (2x )^1 + ^{4}{}{C}_2 (2x )^2 + ^{4}{}{C}_3 (2x )^3 + ^{4}{}{C}_4 (2x )^4 ] + \]
\[ 90 x^4 [1 + 8 x^3 + 6x + 12 x^2 ] - 270 x^6 (1 + 4 x^2 + 4x) + 405 x^8 + 810 x^9 - 243 x^{10} \]
\[ = 1 + 10x + 40 x^2 + 80 x^3 + 80 x^4 + 32 x^5 - 15 x^2 - 120 x^3 - {360}^4 - 480 x^5 - 240 x^6 + \]
\[ 90 x^4 + 720 x^7 + 540 x^5 + 1080 x^6 - 270 x^6 - 1080 x^8 - 1080 x^7 + 405 x^8 + 810 x^9 - 243 x^{10} \]
\[ = 1 + 10x + 25 x^2 - 40 x^3 - 190 x^4 + 92 x^5 + 570 x^6 - 360 x^7 - 675 x^8 + 810 x^9 - 243 x^{10} \]
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