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Question: Expand the given expression \({\left( {1 - 2x} \right)^5}\)...

Expand the given expression (1−2x)5{\left( {1 - 2x} \right)^5}

Explanation

Solution

Hint – In this question use the direct formula for expansion of (1+x)n{\left( {1 + x} \right)^n} according to binomial expansion which is (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+..............{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ............... This will give the answer.

Complete step-by-step solution -
Given equation is
(1−2x)5{\left( {1 - 2x} \right)^5}
Now as we know according to Binomial theorem the expansion of (1+x)n{\left( {1 + x} \right)^n} is
⇒(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+..............\Rightarrow {\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ..............
So the expansion of (1−2x)5{\left( {1 - 2x} \right)^5} according to binomial theorem we have,
⇒(1−2x)5=1+5(−2x)+5(5−1)2!(−2x)2+5(5−1)(5−2)3!(−2x)3+5(5−1)(5−2)(5−3)4!(−2x)4+\Rightarrow {\left( {1 - 2x} \right)^5} = 1 + 5\left( { - 2x} \right) + \dfrac{{5\left( {5 - 1} \right)}}{{2!}}{\left( { - 2x} \right)^2} + \dfrac{{5\left( {5 - 1} \right)\left( {5 - 2} \right)}}{{3!}}{\left( { - 2x} \right)^3} + \dfrac{{5\left( {5 - 1} \right)\left( {5 - 2} \right)\left( {5 - 3} \right)}}{{4!}}{\left( { - 2x} \right)^4} +
+5(5−1)(5−2)(5−3)(5−4)5!(−2x)5+ \dfrac{{5\left( {5 - 1} \right)\left( {5 - 2} \right)\left( {5 - 3} \right)\left( {5 - 4} \right)}}{{5!}}{\left( { - 2x} \right)^5}.
Now simplify the above equation we have,
⇒(1−2x)5=1−10x+5(4)2×1(4x2)+5(4)(3)3×2×1(−8x3)+5(4)(3)(2)4×3×2×1(16x4)+5(4)(3)(2)(1)5×4×3×2×1(−32x5)\Rightarrow {\left( {1 - 2x} \right)^5} = 1 - 10x + \dfrac{{5\left( 4 \right)}}{{2 \times 1}}\left( {4{x^2}} \right) + \dfrac{{5\left( 4 \right)\left( 3 \right)}}{{3 \times 2 \times 1}}\left( { - 8{x^3}} \right) + \dfrac{{5\left( 4 \right)\left( 3 \right)\left( 2 \right)}}{{4 \times 3 \times 2 \times 1}}\left( {16{x^4}} \right) + \dfrac{{5\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}}{{5 \times 4 \times 3 \times 2 \times 1}}\left( { - 32{x^5}} \right)
[∵n! = n(n − 1)(n − 2)(n − 3)(n −4)(n − 5).....................]\left[ {\because n!{\text{ }} = {\text{ }}n\left( {n{\text{ }}-{\text{ }}1} \right)\left( {n{\text{ }}-{\text{ }}2} \right)\left( {n{\text{ }}-{\text{ }}3} \right)\left( {n{\text{ }}-4} \right)\left( {n{\text{ }}-{\text{ }}5} \right).....................} \right] ⇒(1−2x)5=1−10x+40x2−80x3+80x4−32x5 \Rightarrow {\left( {1 - 2x} \right)^5} = 1 - 10x + 40{x^2} - 80{x^3} + 80{x^4} - 32{x^5}
So this is the required expansion of(1−2x)5{\left( {1 - 2x} \right)^5}.

Note – Binomial theorem is one which specifies the expansion of any power (a+b)m{(a + b)^m}of a binomial (a+b)(a + b)as a certain sum of products aibj{a^i}{b^j} such as (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab is also an example a binomial expansion and is derived using this similar concept.