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Question: Find the coefficient of \[{x^7}\] in \[{\left( {1 + x + {x^4} + {x^5}} \right)^6}\]...

Find the coefficient of x7{x^7} in (1+x+x4+x5)6{\left( {1 + x + {x^4} + {x^5}} \right)^6}

Explanation

Solution

In the given question, we have been given a polynomial in a single variable. The polynomial consists of four terms (each term has the same base – the single variable, but different exponential power) in a bracket, and the bracket as a whole raised to an integral power. We have to find the coefficient of a power of the single variable, when the whole of the bracket is expanded; this is going to be solved by using binomial theorem. First, we are going to have to factorize the polynomial and then apply the binomial theorem on it, and solve for the answer.

Formula Used:
We are going to use the formula of binomial theorem, which is:
(x+y)n=∑k=0n(n\k)xn−kyk=∑k=0n(n\k)xkyn−k{\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\\k\end{array} \right)} {x^{n - k}}{y^k} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\\k\end{array} \right)} {x^k}{y^{n - k}}

Complete step by step solution:
The given polynomial is (1+x+x4+x5)6{\left( {1 + x + {x^4} + {x^5}} \right)^6}.
First, we are going to factorize it,
(1+x+x4+x5)=(1(1+x)+x4(1+x))=(1+x)(1+x4)\left( {1 + x + {x^4} + {x^5}} \right) = \left( {1\left( {1 + x} \right) + {x^4}\left( {1 + x} \right)} \right) = \left( {1 + x} \right)\left( {1 + {x^4}} \right)
Now, we have to find the coefficient of x7{x^7} in (1+x)6(1+x4)6{\left( {1 + x} \right)^6}{\left( {1 + {x^4}} \right)^6}.
The first bracket can have the powers of xx from 1to61{\rm{ to 6}}, i.e., x1,x2,x3,x4,x5,x6{x^1},{x^2},{x^3},{x^4},{x^5},{x^6}
And, the second bracket can have the powers of xx in the multiples of 44 with each multiplicand being from 1to61{\rm{ to }}6, i.e., x4,x8,x12,x16,x20,x24{x^4},{x^8},{x^{12}},{x^{16}},{x^{20}},{x^{24}}
We know, am×an=am+n{a^m} \times {a^n} = {a^{m + n}}
So, let us see how we can get to x7{x^7}:
Since all the powers are positive, we are going to have to use the ones which are less than 77 as we can only sum.
Clearly, we can see that only one pair fulfils the criteria:
x3{x^3} from the first bracket and x4{x^4} from the second bracket.
Now, let us expand the two brackets using binomials.
First, let us evaluate (1+x)6{\left( {1 + x} \right)^6}
(1+x)6=1+6x+15x2+20x3+15x4+6x5+x6{\left( {1 + x} \right)^6} = 1 + 6x + 15{x^2} + 20{x^3} + 15{x^4} + 6{x^5} + {x^6}
Now, let us simplify (1+x4)6{\left( {1 + {x^4}} \right)^6}
(1+x4)6=1+6x4+15x8+20x12+15x16+6x20+x24{\left( {1 + {x^4}} \right)^6} = 1 + 6{x^4} + 15{x^8} + 20{x^{12}} + 15{x^{16}} + 6{x^{20}} + {x^{24}}
Now, the coefficient of x4=6{x^4} = 6 and the coefficient of x3=20{x^3} = 20.
After multiplying them both, we get
6x4×20x3=120x76{x^4} \times 20{x^3} = 120{x^7}

Hence, the coefficient is 120120.

Note:
In this question, we first factored the polynomial, so that we could apply the known fact of sum of powers. Then we used the binomial theorem to find the coefficient; here, we did not compute more values as we only had one possible combination for the sum of the powers to be equal to seven. So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. This way, we always have in sight what we have and we need to find.