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 in the expansion of (1+x2)12(1+x12)(1+x24) is
A) 12C6
B) 12C6+2
C) 12C6+4
D) 12C6+6
Solution
We can see that in this question (1+x2)12 is in the form of (x+y)n where x=1 and y=x2. We also know that the binomial expansion of above expression is (x+y)n=nC0+nC1x1+nC2x2+nC3x3+....nCnxn. We will expand the above expression in this form. Then, we will multiply (1+x12)(1+x24) these expressions together. Then we will try to find the coefficient of x24.
Formula used:
(x+y)n=nC0+nC1x1+nC2x2+nC3x3+....nCnxnand nCr=(n−r)!r!n!
Complete step by step answer:
We know that Binomial expansion of (x+y)n=nC0+nC1x1+nC2x2+nC3x3+....nCnxn
We have (1+x2)12(1+x12)(1+x24)
We will expand (1+x2)12 by using Binomial expansion as above and we will multiply these two expressions together (1+x12)(1+x24).
Thus, (12C0+12C1x2+12C2x4+12C3x6+12C4x8+12C5x10+12C6x12+.....12C12x24)(1+x12+x24+x36)
⇒x24(12C0+12C6+12C12) (Here we are finding the coefficient of x24.)
(Here, 12C0=12!×0!12! = 1, we know that 0! is 1 and 12C12=0!×12!12! = 1)
⇒x24(1+12C6+1)
⇒x24(12C6+2)
Thus, the coefficient of x24 is 12C6+2.
Hence, Option B is the correct option.
Note:
Students must know the binomial expansion on (x+y)n. They should also take care while using the formula of the combination when they are solving for 12C0&12C1. While solving this question, students should pick all the coefficients of x24 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, nC1, nC2…., nCn are called binomial coefficients and can represented by C0, C1, C2, …., Cn. The total number of terms in the expansion of (x+y)n are (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.