Question

Find coefficient of ${x^{12}}$ in${(1 + {x^3})^7}{(1 + {n^4})^{12}}{(1 + {x^2})^4}$.

Answer 1

Hint: To find the coefficient,we make use of the binomial theorem which states \[{(1 + x)^n} = 1 + {\,^n}{C_1} + {\,^n}{C_2}{({x^n})^2} + {\,^n}{C_3}{({x^n})^2}..... + {\,^n}{C_n}{({x^n})^n}\]


Complete step by step solution:
Coefficient of\[{x^{12}}\,in\,{(1 + {x^3}]^7}{(1 + {x^4}]^2}{[1 + {x^2}]^4}\]
Expanding using binomial theorem
${(1 + {x^3})^7} = 1 + {\,^7}{C_1}{x^3} + {\,^7}{C_2}{({x^3})^2} + {\,^7}{C_3} + {\,^7}{C_4}{({x^3})^7}$
$ = 1 + \dfrac{{7!}}{{7 - 1!1!}}{x^3} + \dfrac{{7!}}{{7 - 2!2!}}{x^6} + \dfrac{{7!}}{{7 - 33!}}{x^9} + \dfrac{{7!}}{{7 - 4!4!}}{x^{12}} + ......$
\[ = 1 + \dfrac{{7 \times 6!}}{{6!}}{x^3} + \dfrac{{7 \times 6 \times 5!}}{{5! \times 2 \times 1}}{x^6} + \dfrac{{7 \times 6 \times 5 \times 4!}}{{4!3 \times 2 \times 1}}{x^9} + \dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{3!4 \times 3 \times 2 \times 1}}{x^{12}}\]
\[ = 1 + 7{x^3} + 21{x^6} + 35{x^9} + 35{x^{12}}\] ……(1)
Next term${(1 + {x^4}]^2}$,
Expanding using binomial theorem
\[{[1 + {x^4}]^{12}} = 1 + {\,^{12}}{C_1}{x^4} + {\,^{12}}{C_2}{({x^4})^2} + {\,^{12}}{C_3}{({x^4})^3} + ..... + {\,^{12}}{C_{12}}{({x^4})^{12}}\]
$1 + {x^4}{]^{12}} = 1 + \dfrac{{12!}}{{12 - 1!1!}}{x^4} + \dfrac{{12!}}{{12 - 2!2!}}{x^8} + \dfrac{{12!}}{{12 - 3!3!}}{x^{12}} + .....$
\[{[1 + {x^4}]^{12}} = 1 + \dfrac{{12 \times 11!}}{{11!}}{x^4} + \dfrac{{12!}}{{10!2!}}{x^8} + \dfrac{{12!}}{{9!3!}}{x^{12}} + .....\]
${[1 + {x^4}]^{12}} = 1 + 12{x^4} + \dfrac{{12 \times 11 \times 10!}}{{10! \times 2 \times 1}}{x^8} + \dfrac{{12 \times 11 \times 10 \times 9!}}{{3 \times 2 \times 1}}{x^{12}} + ...$
${[1 + {x^4}]^2} = 1 + 12{x^4} + 66{x^8} + 220{x^{12}}$
The last term is ${(1 + {x^2})^4}$
Expanding by binomial theorem, we have
\[{[1 + {x^2}]^4} = 1 + {\,^4}{C_1}{x^2} + {\,^4}{C_2}{({x^2})^2} + {\,^4}{C_3}{({x^2})^3} + {\,^4}{C_4}({x^2})4\]
\[ = 1 + \dfrac{{4!}}{{4 - 1!1!}}{x^2} + \dfrac{{4!}}{{4 - 2!2!}} \times {x^4} + \dfrac{{4!}}{{4 - 3!3!}}{x^6} + \dfrac{{4!}}{{4!0!}}{x^8}\]
\[ = 1 + \dfrac{{4 \times 3!}}{{3!}}{x^2} + \dfrac{{4 \times 3 \times 2!}}{{2!2 \times 1}}{x^4} + \dfrac{{4 \times 3!}}{{1!3!}}{x^6} + 1.{x^8}\]
$ = 1 + 4{x^2} + 6{x^4} + 4{x^6} + {x^8}$
Now, we will multiply the terms${(1 + {x^3})^7}{(1 + {x^4})^{12}}$\[{(1 + {x^2})^4}\], we have
${(1 + {x^3})^7}{(1 + {x^4})^{12}}{(1 + {x^2})^4} = (1 + 7{x^3} + 21{x^6} + 35{x^9} + 35{x^{12}}....)$$(1 + 12{x^4} + 66{x^2} + 220{x^{12}})(1 + 4{x^2} + 6{x^4} + 4{x^6} + {x^8})$
Now, we will equate those values, which will give the coefficient of ${x^{12}}$
\[(1 + {x^3}){(1 + {x^4})^{12}}{(1 + {x^2})^4} = {x^{2l + 3b + 4c}} = {x^{12}}\]
$2l,3b,4c = (0,8,4),(6,4,2),(6,0,6),(12,0,0),(0,12,10)$
$ = 1 \times 66{x^8}.6{x^4} + 21{x^6} \times 12{x^4} \times 4{x^2} + 21{x^6} \times 1 \times 4{x^6} + 35{x^{12}} \times 1 \times 1 + 1 \times 220{x^{12}} \times 1$
Equating the coefficient of ${x^{12}}$
$ = 396 + 1008 + 84 + 35 + 220$
$ = 1734$


Note: Make sure to write the coefficient which is associated with the variable which is asked.Do not consider other variables when writing the coefficient asked