Solveeit Logo

Question

Question: The coefficient of \[{x^{24}}\] in the expansion of \({(1 + {x^2})^{12}}\,(1 + {x^{12}})\,(1 + {x^{2...

The coefficient of x24{x^{24}} in the expansion of (1+x2)12(1+x12)(1+x24){(1 + {x^2})^{12}}\,(1 + {x^{12}})\,(1 + {x^{24}}) is
A) 12C6^{12}{C_6}
B) 12C6+2^{12}{C_6} + 2
C) 12C6+4^{12}{C_6} + 4
D) 12C6+6^{12}{C_6} + 6

Explanation

Solution

We can see that in this question (1+x2)12{(1 + {x^2})^{12}} is in the form of (x+y)n{(x + y)^n} where x=1x = 1 and y=x2y={x^2}. We also know that the binomial expansion of above expression is (x+y)n=nC0+nC1x1+nC2x2+nC3x3+....nCnxn{(x + y)^n}{ = ^n}{C_0}{ + ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2}{ + ^n}{C_3}{x^3} + ....{\,^n}{C_n}{x^n}. We will expand the above expression in this form. Then, we will multiply (1+x12)(1+x24)(1 + {x^{12}})\,(1 + {x^{24}}) these expressions together. Then we will try to find the coefficient of x24{x^{24}}.

Formula used:
(x+y)n=nC0+nC1x1+nC2x2+nC3x3+....nCnxn{(x + y)^n}{ = ^n}{C_0}{ + ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2}{ + ^n}{C_3}{x^3} + ....{\,^n}{C_n}{x^n}and nCr=n!(nr)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}

Complete step by step answer:
We know that Binomial expansion of (x+y)n=nC0+nC1x1+nC2x2+nC3x3+....nCnxn{(x + y)^n}{ = ^n}{C_0}{ + ^n}{C_1}{x^1}{ + ^n}{C_2}{x^2}{ + ^n}{C_3}{x^3} + ....{\,^n}{C_n}{x^n}
We have (1+x2)12(1+x12)(1+x24){(1 + {x^2})^{12}}\,(1 + {x^{12}})\,(1 + {x^{24}})
We will expand (1+x2)12{(1 + {x^2})^{12}} by using Binomial expansion as above and we will multiply these two expressions together (1+x12)(1+x24)(1 + {x^{12}})\,(1 + {x^{24}}).
Thus, (12C0+12C1x2+12C2x4+12C3x6+12C4x8+12C5x10+12C6x12+.....12C12x24)(1+x12+x24+x36){(^{12}}{C_0}{ + ^{12}}{C_1}{x^2}{ + ^{12}}{C_2}{x^4}{ + ^{12}}{C_3}{x^6}{ + ^{12}}{C_4}{x^8}{ + ^{12}}{C_5}{x^{10}}{ + ^{12}}{C_6}{x^{12}} + {.....^{12}}{C_{12}}{x^{24}})\,(1 + {x^{12}} + {x^{24}} + {x^{36}})
x24(12C0+12C6+12C12)\Rightarrow {x^{24}}{(^{12}}{C_0}{ + ^{12}}{C_6}{ + ^{12}}{C_{12}}) (Here we are finding the coefficient of x24{x^{24}}.)
(Here, 12C0=12!12!×0!^{12}{C_0} = \dfrac{{12!}}{{12! \times 0!}} = 1, we know that 0! is 1 and 12C12=12!0!×12!^{12}{C_{12}} = \dfrac{{12!}}{{0! \times 12!}} = 1)
x24(1+12C6+1)\Rightarrow {x^{24}}(1{ + ^{12}}{C_6} + 1)
x24(12C6+2)\Rightarrow {x^{24}}{(^{12}}{C_6} + 2)
Thus, the coefficient of x24{x^{24}} is 12C6+2^{12}{C_6} + 2.

Hence, Option B is the correct option.

Note:
Students must know the binomial expansion on (x+y)n{(x + y)^n}. They should also take care while using the formula of the combination when they are solving for 12C0^{12}{C_0}&12C1^{12}{C_1}. While solving this question, students should pick all the coefficients of x24{x^{24}} carefully. If any of the coefficients is left then you will get an incorrect answer. You might find Binomial expansion lengthy and tedious to calculate. But a binomial expression that has large power can be easily calculated with the help of the Binomial Theorem. nC0^n{C_0}, nC1^n{C_1}, nC2^n{C_2}…., nCn^n{C_n} are called binomial coefficients and can represented by C0{C_0}, C1{C_1}, C2{C_2}, …., Cn{C_n}. The total number of terms in the expansion of (x+y)n{\left( {x + y} \right)^n} are (n+1)(n + 1). These are a few important things about binomial expansion. You should keep all these things in your mind while solving these types of questions.