A) 52
B) 44
C) 50
D) 56
Correct Answer: A
Solution :
\[{{(1+x)}^{2}}=1+2x+{{x}^{2}}\] \[{{(1+{{x}^{2}})}^{3}}=1+3{{x}^{2}}+3{{x}^{4}}+{{x}^{6}}\] \[{{(1={{x}^{3}})}^{4}}=1+4{{x}^{3}}+6{{x}^{6}}+4{{x}^{9}}+{{x}^{12}}\] From the above the binomial expansion, we want the terms containing \[{{x}^{10}}\] after multiplication So, the combinations are: \[x.{{x}^{9}},x.{{x}^{6}}.{{x}^{3}},{{x}^{2}}.{{x}^{2}}.{{x}^{6}},{{x}^{4}}.{{x}^{6}}\] Their coefficients are\[2\times 4,2\times 1\times 4,1\times 3\times 6,3\times 6\] Which is\[8,8,18,18\] Sum of the coefficients is\[8+8+18+18=52\] Hence, the coefficient of \[{{x}^{10}}\] in the expansion of \[{{(1+x)}^{2}}{{(1+{{x}^{2}})}^{3}}{{(1+{{x}^{3}})}^{4}}\] is 52 To find: Coefficient of \[{{x}^{10}}\] Given, \[{{(1+x)}^{2}}{{(1+{{x}^{2}})}^{3}}{{(1+{{x}^{3}})}^{4}}\] Using the above series using binomial coefficients, \[\Rightarrow \left[ \left( _{0}^{2} \right){{x}^{2}}+\left( _{1}^{2} \right)x+\left( _{2}^{2} \right) \right]\times \left[ \left( _{0}^{3} \right){{x}^{6}}+\left( _{1}^{3} \right){{x}^{4}}+\left( _{2}^{3} \right){{x}^{2}}+\left( _{3}^{3} \right) \right]\times \] \[\left[ \left( _{0}^{4} \right){{x}^{12}}+\left( _{1}^{4} \right){{x}^{9}}+\left( _{2}^{4} \right){{x}^{9}}+\left( _{2}^{4} \right){{x}^{6}}+\left( _{3}^{4} \right){{x}^{3}}+\left( _{4}^{4} \right) \right]\] Finding the coefficient of\[{{x}^{10}}\]: Multiply individual term to obtain the term \[{{x}^{10}}\] and note down its coefficients. Find all such possible combinations. 1. \[\left( _{1}^{4} \right){{x}^{9}}\times \left( _{3}^{3} \right)\times \left( _{1}^{2} \right)x=\left( _{1}^{4} \right)\times \left( _{3}^{3} \right)\times \left( _{1}^{2} \right){{x}^{10}}=8{{x}^{10}}\] 2. \[\left( _{1}^{4} \right){{x}^{6}}\times \left( _{1}^{3} \right){{x}^{4}}\times \left( _{2}^{2} \right)=\left( _{2}^{4} \right)\times \left( _{1}^{3} \right)\times \left( _{2}^{2} \right){{x}^{10}}=18{{x}^{10}}\] 3. \[\left( _{2}^{4} \right){{x}^{6}}\times \left( _{2}^{3} \right){{x}^{2}}\times \left( _{0}^{2} \right){{x}^{2}}=\left( _{2}^{4} \right)\times \left( _{2}^{3} \right)\times \left( _{0}^{2} \right){{x}^{10}}=18{{x}^{10}}\] 4. \[\left( _{3}^{4} \right){{x}^{3}}\times \left( _{0}^{3} \right){{x}^{6}}\times \left( _{1}^{2} \right)x=\left( _{3}^{4} \right)\times \left( _{0}^{3} \right)\times \left( _{1}^{2} \right){{x}^{10}}=8{{x}^{10}}\] Adding all the coefficients of \[{{x}^{10}}\] we get, \[52{{x}^{10}}\]. Therefore, the coefficient of \[{{x}^{10}}\] in the expansion is \[52\].
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