Using Binomial Theorem, the expression (1 2x)^5 can be expanded as nbsp; nbsp; (1 2x)^5= ^5C_0(1)^5 ^5C_1(1)^4(2x)+^5C_2(1)^3(2x)^2 ^5C_3(1)^2(2x) ...

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Byju's Answer

Standard XI

Mathematics

Sum of product of binomial coefficients

Expand 1-2x...

Question

Expand $(1 - 2 x)^{5}$

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Solution

Using Binomial Theorem, the expression $(1 - 2 x)^{5}$ can be expanded as
$(1 - 2 x)^{5} =^{5} C_{0} (1)^{5} -^{5} C_{1} (1)^{4} (2 x) +^{5} C_{2} (1)^{3} (2 x)^{2} -^{5} C_{3} (1)^{2} (2 x)^{3} +^{5} C_{4} (1)^{1} (2 x)^{4} -^{5} C_{5} (2 x)^{5}$
$= 1 - 5 (2 x) + 10 (4 x^{2}) - 10 (8 x^{3}) + 5 (16 x^{4}) - (32 x^{5})$
$= 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5}$

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Similar questions

Expand the expression (1– 2x)⁵

Q. Expand each of the expressions in 1 to 2.

(1 - 2 x)^{5}

{(\frac{2}{x} - \frac{x}{2})}^{5}

(1– 2x)⁵

\int \frac{x^{7} + 2 x^{5} + x^{3} + 1}{x^{2} + 1} d x =

${lim}_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}$

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Expand (1−2x)5

Using Binomial Theorem, the expression (1−2x)5 can be expanded as (1−2x)5=5C0(1)5−5C1(1)4(2x)+5C2(1)3(2x)2−5C3(1)2(2x)3+5C4(1)1(2x)4−5C5(2x)5=1−5(2x)+10(4x2)−10(8x3)+5(16x4)−(32x5)=1−10x+40x2−80x3+80x4−32x5

Expand $(1 - 2 x)^{5}$