A) \[{{4}^{n}}\]
B) 1
C) -1
D) 0
Correct Answer: D
Solution :
Given, \[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{n}}=\sum\limits_{r=0}^{3n}{{{a}_{r}}{{x}^{r}}}\]and \[n\]is an odd positive integer. \[\Rightarrow \]\[{{[(1+x){{(1+x)}^{2}}]}^{n}}=\sum\limits_{r=0}^{3n}{{{a}_{r}}{{x}^{{}}}}\] \[\Rightarrow \]\[{{(1+x)}^{n}}{{(1+{{x}^{2}})}^{n}}=\sum\limits_{r=0}^{3n}{{{a}_{r}}{{x}^{r}}}\] If we take \[n=1,\] \[(1+x+{{x}^{2}}+{{x}^{3}})=\sum\limits_{r=0}^{3}{{{a}_{r}}{{x}^{r}}}\] \[{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}\] On comparing both sides, \[{{a}_{0}}=1,{{a}_{1}}=1,{{a}_{2}}=1,{{a}_{3}}=1\] ?(i) If we take \[n=3,\] \[{{(1+x)}^{3}}{{(1+{{x}^{2}})}^{3}}=\sum\limits_{r=0}^{9}{{{a}_{r}}{{x}^{r}}}\] \[(1+{{x}^{3}}+3{{x}^{2}}+3x)(1+{{x}^{6}}+3{{x}^{4}}+3{{x}^{2}})\] \[\sum\limits_{r=0}^{9}{{{a}_{r}}{{x}^{r}}(1+{{x}^{3}}+3{{x}^{2}}+3x+{{x}^{6}}+{{x}^{9}}}\] \[+3{{x}^{8}}+3{{x}^{7}}+3{{x}^{4}}+3{{x}^{7}}+9{{x}^{6}}\] \[+9{{x}^{5}}+3{{x}^{2}}+3{{x}^{5}}+9{{x}^{4}}+9{{x}^{3}})\] \[=\sum\limits_{r=0}^{9}{{{a}_{r}}{{x}^{r}}(1+3x+6{{x}^{2}}+10{{x}^{3}}+12{{x}^{4}}}\] \[+12{{x}^{5}}+10{{x}^{6}}+6{{x}^{7}}+3{{x}^{8}}+{{x}^{9}})\] \[=\sum\limits_{r=0}^{9}{{{a}_{r}}{{x}^{r}}}\] On comparing the coefficient of\[x\]on both sides; \[{{a}_{0}}=1,\,{{a}_{1}}=3,\,{{a}_{2}}=6,\,{{a}_{3}}=10,\,{{a}_{4}}=12,\,{{a}_{5}}=12,\] \[{{a}_{6}}=10,\,{{a}_{7}}=6,\,{{a}_{8}}=3,\,{{a}_{9}}=1\] ?(ii) From Eq.(i), we see that, \[{{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}=0,\]when \[n=1\] From Eq. (ii), we see that, \[{{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+{{a}_{4}}-{{a}_{5}}+{{a}_{6}}-{{a}_{7}}\] \[+{{a}_{8}}-{{a}_{9}}=0\] when \[n=3,\] similarly, for each odd terms: \[{{a}_{0}}-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+...-{{a}_{3n}}=0\]
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