Expand using binomial theorem \[{{\left( 1+\dfrac{x}{2}-\dfrac{2}{x} \right)}^{4}}\], \[x\ne 0\] and let the sum of coefficient of the terms in the expansion be t. Find 10000t.
Answer
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Hint: In the above question, we will use the formula of binomial expansion to expand the given expression and the formula is as follows:
\[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...{{+}^{n}}{{C}_{n}}{{b}^{n}}\]
We will use the above formula for n = 4 and thus we will get all the coefficient of the terms in the expansion.
Complete step-by-step answer:
We know that \[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...{{+}^{n}}{{C}_{n}}{{b}^{n}}\]. For n = 4, we as follows:
\[\begin{align}
& {{\left( a+b \right)}^{4}}{{=}^{4}}{{C}_{0}}{{a}^{4}}{{+}^{4}}{{C}_{1}}{{a}^{3}}{{b}^{1}}{{+}^{4}}{{C}_{2}}{{a}^{2}}{{b}^{2}}{{+}^{4}}{{C}_{3}}{{a}^{1}}{{b}^{3}}{{+}^{4}}{{C}_{4}}{{b}^{4}} \\
& =\dfrac{4!}{0!\left( 4-0 \right)!}{{a}^{4}}+\dfrac{4!}{1\times \left( 4-1 \right)!}{{a}^{3}}{{b}^{1}}+\dfrac{4!}{2!\left( 4-2 \right)!}{{a}^{2}}{{b}^{2}}+\dfrac{4!}{3!\left( 4-3 \right)!}{{a}^{1}}{{b}^{3}}+\dfrac{4!}{4!\left( 4-4 \right)!}{{b}^{4}} \\
& =\dfrac{4!}{1\times 4!}{{a}^{4}}+\dfrac{4!}{1\times 3!}{{a}^{3}}{{b}^{1}}+\dfrac{4!}{2!\times 2!}{{a}^{2}}{{b}^{2}}+\dfrac{4!}{3!\times 1}{{a}^{1}}{{b}^{3}}+\dfrac{4!}{4!0!}{{b}^{4}} \\
& ={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}.....(1) \\
\end{align}\]
Hence, \[{{\left( a+b \right)}^{4}}={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}\].
We have been given the expression \[{{\left( 1+\dfrac{x}{2}-\dfrac{2}{x} \right)}^{4}}\].
On substituting \[a=\left( 1+\dfrac{x}{2} \right)\] and \[b=\left( -\dfrac{2}{x} \right)\], we get as follows:
\[\begin{align}
& {{\left( 1+\dfrac{x}{2}-\dfrac{2}{x} \right)}^{4}}={{\left( 1+\dfrac{x}{2} \right)}^{4}}+4{{\left( 1+\dfrac{x}{2} \right)}^{3}}\left( \dfrac{-2}{x} \right)+6{{\left( 1+\dfrac{x}{2} \right)}^{2}}{{\left( \dfrac{-2}{x} \right)}^{2}}+4\left( 1+\dfrac{x}{2} \right){{\left( \dfrac{-2}{x} \right)}^{3}}+{{\left( \dfrac{-2}{x} \right)}^{4}} \\
& ={{\left( 1+\dfrac{x}{2} \right)}^{4}}-\dfrac{8}{x}{{\left( 1+\dfrac{x}{2} \right)}^{3}}+\dfrac{24}{{{x}^{2}}}{{\left( 1+\dfrac{x}{2} \right)}^{2}}-\dfrac{32}{{{x}^{3}}}\left( 1+\dfrac{x}{2} \right)+\dfrac{16}{{{x}^{4}}} \\
\end{align}\]
Now we will solve \[{{\left( 1+\dfrac{x}{2} \right)}^{4}}\] and \[{{\left( 1+\dfrac{x}{2} \right)}^{3}}\] separately.
We have \[{{\left( 1+\dfrac{x}{2} \right)}^{4}}\].
On using the equation (1) and substituting a = 1 and b = \[\dfrac{x}{4}\], we get as follows:
\[\begin{align}
& {{\left( 1+\dfrac{x}{2} \right)}^{4}}={{\left( 1 \right)}^{4}}+4{{\left( 1 \right)}^{3}}\left( \dfrac{x}{2} \right)+6{{\left( 1 \right)}^{2}}{{\left( \dfrac{x}{2} \right)}^{2}}+4\left( 1 \right){{\left( \dfrac{x}{2} \right)}^{3}}+{{\left( \dfrac{x}{2} \right)}^{4}} \\
& =1+4\left( \dfrac{x}{2} \right)+6\left( \dfrac{{{x}^{2}}}{4} \right)+4\left( \dfrac{{{x}^{3}}}{8} \right)+\left( \dfrac{{{x}^{4}}}{16} \right) \\
& =1+2x+\dfrac{3}{2}{{x}^{2}}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16} \\
\end{align}\]
Now we have \[{{\left( 1+\dfrac{x}{2} \right)}^{3}}\].
As we know that \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]
\[\begin{align}
& {{\left( 1+\dfrac{x}{2} \right)}^{3}}={{1}^{3}}+{{\left( \dfrac{x}{2} \right)}^{3}}+3\times 1\times \dfrac{x}{2}\left( 1+\dfrac{x}{2} \right) \\
& =1+\dfrac{{{x}^{3}}}{8}+\dfrac{3x}{2}+\dfrac{3{{x}^{2}}}{4} \\
& =1+\dfrac{3x}{2}+\dfrac{3{{x}^{2}}}{4}+\dfrac{{{x}^{3}}}{8} \\
\end{align}\]
On substituting the values of \[{{\left( 1+\dfrac{x}{2} \right)}^{4}}\] and \[{{\left( 1+\dfrac{x}{2} \right)}^{3}}\], we get as follows:
\[\begin{align}
& {{\left[ \left( 1+\dfrac{x}{2} \right)-\dfrac{2}{x} \right]}^{4}}=\left( 1+2x+\dfrac{3{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16} \right)-\dfrac{8}{x}\left( 1+\dfrac{3x}{2}+\dfrac{3{{x}^{2}}}{4}+\dfrac{{{x}^{3}}}{8} \right)+\dfrac{24}{{{x}^{2}}}\left( 1+{{\left( \dfrac{x}{2} \right)}^{2}}+2\times 1\times \dfrac{x}{2} \right) \\
& -\dfrac{32}{{{x}^{3}}}\left( 1+\dfrac{x}{2} \right)+\dfrac{16}{{{x}^{4}}} \\
& =\left( 1+2x+\dfrac{3{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16} \right)-\left( \dfrac{8}{x}+12+6x+{{x}^{2}} \right)+\left( \dfrac{24}{{{x}^{2}}}+\dfrac{24}{{{x}^{2}}}\times \dfrac{{{x}^{2}}}{4}+\dfrac{24}{{{x}^{2}}}\times x \right)-\left( \dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{2}}} \right)+\dfrac{16}{{{x}^{4}}} \\
& =1+2x+\dfrac{3{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16}-\dfrac{8}{x}-12-6x-{{x}^{2}}+\dfrac{24}{{{x}^{2}}}+6+\dfrac{24}{x}-\dfrac{32}{{{x}^{3}}}-\dfrac{16}{{{x}^{2}}}+\dfrac{16}{{{x}^{4}}} \\
& =\dfrac{{{x}^{4}}}{16}+\dfrac{{{x}^{3}}}{2}+\dfrac{3{{x}^{2}}}{2}-{{x}^{2}}+2x-6x+1-12+6-\dfrac{8}{x}+\dfrac{24}{x}+\dfrac{24}{{{x}^{2}}}-\dfrac{16}{{{x}^{2}}}-\dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{4}}} \\
& =\dfrac{{{x}^{4}}}{16}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{2}}}{2}-4x-5+\dfrac{16}{x}+\dfrac{8}{{{x}^{2}}}-\dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{4}}} \\
\end{align}\]
Thus, \[{{\left[ \left( 1+\dfrac{x}{2} \right)-\dfrac{2}{x} \right]}^{4}}=\dfrac{{{x}^{4}}}{16}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{2}}}{2}-4x-5+\dfrac{16}{x}+\dfrac{8}{{{x}^{2}}}-\dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{4}}}\].
The coefficient of the terms are \[\dfrac{1}{16},\dfrac{1}{2},-4,-5,16,8,-32,16\].
\[\begin{align}
& t=\dfrac{1}{16}+\dfrac{1}{2}+\dfrac{1}{2}-4-5+16+8-32+16 \\
& t=\dfrac{1}{16}+1-9+16+16-24 \\
& t=\dfrac{1}{16}-8+8 \\
& t=\dfrac{1}{16} \\
\end{align}\]
So, \[10000t=10000\times \dfrac{1}{16}=625\].
Therefore, the value of 10000t equals 625.
Note: Take care of the sign and be careful at each and every step of calculation as there is a chance that you might make a mistake with the signs. Also, remember the point that in competitive examination, if we are asked to find the sum of coefficients of the expression given in the question then we can get the answer by substituting x = 1 in the expression.
\[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...{{+}^{n}}{{C}_{n}}{{b}^{n}}\]
We will use the above formula for n = 4 and thus we will get all the coefficient of the terms in the expansion.
Complete step-by-step answer:
We know that \[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...{{+}^{n}}{{C}_{n}}{{b}^{n}}\]. For n = 4, we as follows:
\[\begin{align}
& {{\left( a+b \right)}^{4}}{{=}^{4}}{{C}_{0}}{{a}^{4}}{{+}^{4}}{{C}_{1}}{{a}^{3}}{{b}^{1}}{{+}^{4}}{{C}_{2}}{{a}^{2}}{{b}^{2}}{{+}^{4}}{{C}_{3}}{{a}^{1}}{{b}^{3}}{{+}^{4}}{{C}_{4}}{{b}^{4}} \\
& =\dfrac{4!}{0!\left( 4-0 \right)!}{{a}^{4}}+\dfrac{4!}{1\times \left( 4-1 \right)!}{{a}^{3}}{{b}^{1}}+\dfrac{4!}{2!\left( 4-2 \right)!}{{a}^{2}}{{b}^{2}}+\dfrac{4!}{3!\left( 4-3 \right)!}{{a}^{1}}{{b}^{3}}+\dfrac{4!}{4!\left( 4-4 \right)!}{{b}^{4}} \\
& =\dfrac{4!}{1\times 4!}{{a}^{4}}+\dfrac{4!}{1\times 3!}{{a}^{3}}{{b}^{1}}+\dfrac{4!}{2!\times 2!}{{a}^{2}}{{b}^{2}}+\dfrac{4!}{3!\times 1}{{a}^{1}}{{b}^{3}}+\dfrac{4!}{4!0!}{{b}^{4}} \\
& ={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}.....(1) \\
\end{align}\]
Hence, \[{{\left( a+b \right)}^{4}}={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}\].
We have been given the expression \[{{\left( 1+\dfrac{x}{2}-\dfrac{2}{x} \right)}^{4}}\].
On substituting \[a=\left( 1+\dfrac{x}{2} \right)\] and \[b=\left( -\dfrac{2}{x} \right)\], we get as follows:
\[\begin{align}
& {{\left( 1+\dfrac{x}{2}-\dfrac{2}{x} \right)}^{4}}={{\left( 1+\dfrac{x}{2} \right)}^{4}}+4{{\left( 1+\dfrac{x}{2} \right)}^{3}}\left( \dfrac{-2}{x} \right)+6{{\left( 1+\dfrac{x}{2} \right)}^{2}}{{\left( \dfrac{-2}{x} \right)}^{2}}+4\left( 1+\dfrac{x}{2} \right){{\left( \dfrac{-2}{x} \right)}^{3}}+{{\left( \dfrac{-2}{x} \right)}^{4}} \\
& ={{\left( 1+\dfrac{x}{2} \right)}^{4}}-\dfrac{8}{x}{{\left( 1+\dfrac{x}{2} \right)}^{3}}+\dfrac{24}{{{x}^{2}}}{{\left( 1+\dfrac{x}{2} \right)}^{2}}-\dfrac{32}{{{x}^{3}}}\left( 1+\dfrac{x}{2} \right)+\dfrac{16}{{{x}^{4}}} \\
\end{align}\]
Now we will solve \[{{\left( 1+\dfrac{x}{2} \right)}^{4}}\] and \[{{\left( 1+\dfrac{x}{2} \right)}^{3}}\] separately.
We have \[{{\left( 1+\dfrac{x}{2} \right)}^{4}}\].
On using the equation (1) and substituting a = 1 and b = \[\dfrac{x}{4}\], we get as follows:
\[\begin{align}
& {{\left( 1+\dfrac{x}{2} \right)}^{4}}={{\left( 1 \right)}^{4}}+4{{\left( 1 \right)}^{3}}\left( \dfrac{x}{2} \right)+6{{\left( 1 \right)}^{2}}{{\left( \dfrac{x}{2} \right)}^{2}}+4\left( 1 \right){{\left( \dfrac{x}{2} \right)}^{3}}+{{\left( \dfrac{x}{2} \right)}^{4}} \\
& =1+4\left( \dfrac{x}{2} \right)+6\left( \dfrac{{{x}^{2}}}{4} \right)+4\left( \dfrac{{{x}^{3}}}{8} \right)+\left( \dfrac{{{x}^{4}}}{16} \right) \\
& =1+2x+\dfrac{3}{2}{{x}^{2}}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16} \\
\end{align}\]
Now we have \[{{\left( 1+\dfrac{x}{2} \right)}^{3}}\].
As we know that \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]
\[\begin{align}
& {{\left( 1+\dfrac{x}{2} \right)}^{3}}={{1}^{3}}+{{\left( \dfrac{x}{2} \right)}^{3}}+3\times 1\times \dfrac{x}{2}\left( 1+\dfrac{x}{2} \right) \\
& =1+\dfrac{{{x}^{3}}}{8}+\dfrac{3x}{2}+\dfrac{3{{x}^{2}}}{4} \\
& =1+\dfrac{3x}{2}+\dfrac{3{{x}^{2}}}{4}+\dfrac{{{x}^{3}}}{8} \\
\end{align}\]
On substituting the values of \[{{\left( 1+\dfrac{x}{2} \right)}^{4}}\] and \[{{\left( 1+\dfrac{x}{2} \right)}^{3}}\], we get as follows:
\[\begin{align}
& {{\left[ \left( 1+\dfrac{x}{2} \right)-\dfrac{2}{x} \right]}^{4}}=\left( 1+2x+\dfrac{3{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16} \right)-\dfrac{8}{x}\left( 1+\dfrac{3x}{2}+\dfrac{3{{x}^{2}}}{4}+\dfrac{{{x}^{3}}}{8} \right)+\dfrac{24}{{{x}^{2}}}\left( 1+{{\left( \dfrac{x}{2} \right)}^{2}}+2\times 1\times \dfrac{x}{2} \right) \\
& -\dfrac{32}{{{x}^{3}}}\left( 1+\dfrac{x}{2} \right)+\dfrac{16}{{{x}^{4}}} \\
& =\left( 1+2x+\dfrac{3{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16} \right)-\left( \dfrac{8}{x}+12+6x+{{x}^{2}} \right)+\left( \dfrac{24}{{{x}^{2}}}+\dfrac{24}{{{x}^{2}}}\times \dfrac{{{x}^{2}}}{4}+\dfrac{24}{{{x}^{2}}}\times x \right)-\left( \dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{2}}} \right)+\dfrac{16}{{{x}^{4}}} \\
& =1+2x+\dfrac{3{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{16}-\dfrac{8}{x}-12-6x-{{x}^{2}}+\dfrac{24}{{{x}^{2}}}+6+\dfrac{24}{x}-\dfrac{32}{{{x}^{3}}}-\dfrac{16}{{{x}^{2}}}+\dfrac{16}{{{x}^{4}}} \\
& =\dfrac{{{x}^{4}}}{16}+\dfrac{{{x}^{3}}}{2}+\dfrac{3{{x}^{2}}}{2}-{{x}^{2}}+2x-6x+1-12+6-\dfrac{8}{x}+\dfrac{24}{x}+\dfrac{24}{{{x}^{2}}}-\dfrac{16}{{{x}^{2}}}-\dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{4}}} \\
& =\dfrac{{{x}^{4}}}{16}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{2}}}{2}-4x-5+\dfrac{16}{x}+\dfrac{8}{{{x}^{2}}}-\dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{4}}} \\
\end{align}\]
Thus, \[{{\left[ \left( 1+\dfrac{x}{2} \right)-\dfrac{2}{x} \right]}^{4}}=\dfrac{{{x}^{4}}}{16}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{2}}}{2}-4x-5+\dfrac{16}{x}+\dfrac{8}{{{x}^{2}}}-\dfrac{32}{{{x}^{3}}}+\dfrac{16}{{{x}^{4}}}\].
The coefficient of the terms are \[\dfrac{1}{16},\dfrac{1}{2},-4,-5,16,8,-32,16\].
\[\begin{align}
& t=\dfrac{1}{16}+\dfrac{1}{2}+\dfrac{1}{2}-4-5+16+8-32+16 \\
& t=\dfrac{1}{16}+1-9+16+16-24 \\
& t=\dfrac{1}{16}-8+8 \\
& t=\dfrac{1}{16} \\
\end{align}\]
So, \[10000t=10000\times \dfrac{1}{16}=625\].
Therefore, the value of 10000t equals 625.
Note: Take care of the sign and be careful at each and every step of calculation as there is a chance that you might make a mistake with the signs. Also, remember the point that in competitive examination, if we are asked to find the sum of coefficients of the expression given in the question then we can get the answer by substituting x = 1 in the expression.
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