How do you use the binomial theorem to expand \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] ?
Answer
551.4k+ views
Hint: The given question requires us to find the binomial expansion of the given binomial expression \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] . We can find the required binomial expansion by using the Binomial theorem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.
Complete step-by-step answer:
We have to find the binomial expansion of \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] . So, using the binomial theorem, the binomial expansion of \[{\left( {1 + x} \right)^n}\] is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + ....$
So, the binomial expansion of \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] can be calculated by the above expression by substituting in the correct value of n and x.
Now, we compare the expression \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] with \[{\left( {1 + x} \right)^n}\] . So, we get the value of n as $\left( { - \dfrac{1}{2}} \right)$ and $ - x$ in place of $x$.
So, we have, \[ = 1 + \left( { - \dfrac{1}{2}} \right)\left( { - x} \right) + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 1} \right)}}{{2!}}{\left( { - x} \right)^2} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 1} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 2} \right)}}{{3!}}{\left( { - x} \right)^3} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 1} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 2} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 3} \right)}}{{4!}}{\left( { - x} \right)^4} + ....\]
So, simplifying the expression further, we get,
\[ = 1 + \left( { - \dfrac{1}{2}} \right)\left( { - x} \right) + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{2!}}{\left( { - x} \right)^2} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)}}{{3!}}{\left( { - x} \right)^3} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)\left( { - \dfrac{7}{2}} \right)}}{{4!}}{\left( { - x} \right)^4} + ....\]
So, we can evaluate first few terms of the binomial expansion,
\[ = 1 + \dfrac{1}{2}x + \dfrac{{\left( {\dfrac{3}{4}} \right)}}{2}{x^2} + \dfrac{{\left( {\dfrac{{ - 15}}{8}} \right)}}{6}\left( { - {x^3}} \right) + \dfrac{{\left( {\dfrac{{105}}{{16}}} \right)}}{{24}}{x^4} + ....\]
Simplifying the expression and opening the brackets, we get,
\[ = 1 + \dfrac{1}{2}x + \dfrac{3}{8}{x^2} + \dfrac{5}{{16}}{x^3} + \dfrac{{35}}{{128}}{x^4} + ....\]
Hence, the expression \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] can be simplified as \[\left( {1 + \dfrac{1}{2}x + \dfrac{3}{8}{x^2} + \dfrac{5}{{16}}{x^3} + \dfrac{{35}}{{128}}{x^4} + ....} \right)\] using binomial theorem.
So, the correct answer is “ \[\left( {1 + \dfrac{1}{2}x + \dfrac{3}{8}{x^2} + \dfrac{5}{{16}}{x^3} + \dfrac{{35}}{{128}}{x^4} + ....} \right)\] ”.
Note: The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. One should be careful while carrying out the calculations and simplification.
Complete step-by-step answer:
We have to find the binomial expansion of \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] . So, using the binomial theorem, the binomial expansion of \[{\left( {1 + x} \right)^n}\] is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + ....$
So, the binomial expansion of \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] can be calculated by the above expression by substituting in the correct value of n and x.
Now, we compare the expression \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] with \[{\left( {1 + x} \right)^n}\] . So, we get the value of n as $\left( { - \dfrac{1}{2}} \right)$ and $ - x$ in place of $x$.
So, we have, \[ = 1 + \left( { - \dfrac{1}{2}} \right)\left( { - x} \right) + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 1} \right)}}{{2!}}{\left( { - x} \right)^2} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 1} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 2} \right)}}{{3!}}{\left( { - x} \right)^3} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 1} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 2} \right)\left( {\left( { - \dfrac{1}{2}} \right) - 3} \right)}}{{4!}}{\left( { - x} \right)^4} + ....\]
So, simplifying the expression further, we get,
\[ = 1 + \left( { - \dfrac{1}{2}} \right)\left( { - x} \right) + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{2!}}{\left( { - x} \right)^2} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)}}{{3!}}{\left( { - x} \right)^3} + \dfrac{{\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)\left( { - \dfrac{7}{2}} \right)}}{{4!}}{\left( { - x} \right)^4} + ....\]
So, we can evaluate first few terms of the binomial expansion,
\[ = 1 + \dfrac{1}{2}x + \dfrac{{\left( {\dfrac{3}{4}} \right)}}{2}{x^2} + \dfrac{{\left( {\dfrac{{ - 15}}{8}} \right)}}{6}\left( { - {x^3}} \right) + \dfrac{{\left( {\dfrac{{105}}{{16}}} \right)}}{{24}}{x^4} + ....\]
Simplifying the expression and opening the brackets, we get,
\[ = 1 + \dfrac{1}{2}x + \dfrac{3}{8}{x^2} + \dfrac{5}{{16}}{x^3} + \dfrac{{35}}{{128}}{x^4} + ....\]
Hence, the expression \[{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\] can be simplified as \[\left( {1 + \dfrac{1}{2}x + \dfrac{3}{8}{x^2} + \dfrac{5}{{16}}{x^3} + \dfrac{{35}}{{128}}{x^4} + ....} \right)\] using binomial theorem.
So, the correct answer is “ \[\left( {1 + \dfrac{1}{2}x + \dfrac{3}{8}{x^2} + \dfrac{5}{{16}}{x^3} + \dfrac{{35}}{{128}}{x^4} + ....} \right)\] ”.
Note: The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. One should be careful while carrying out the calculations and simplification.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

