Question 245 of 347MediumMCQ
The coefficient of x−5 in the binomial expansion of (x2/3−x1/3+1x+1−x−x1/2x−1)10 where x=0,1, is:
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Official solution
\[{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}\]
= \[{\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}\]
= \[{\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}\]
= \[{\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}\]
= \[{\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}\]
= \[{\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}\]
[Note:
For \[{\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}\] the \[\left( {r + 1} \right)\]th term with power m of x is
\[r = {{n\alpha - m} \over {\alpha + \beta }}\]]
Here \[\alpha = {1 \over 3}\], \[\beta = {1 \over 2}\] and m = -5
then \[r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}\] = \[{{25} \over 3} \times {6 \over 5}\] = 10
\[\therefore\] T11 is the term with x-5.
\[\therefore\] T11 = 10C10 = 1
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