Coefficient of a Sixteenth Power

The problem of finding the coefficient of $x^{16}$ is a mater of finding all the way one can chose $x^{16}$ out of 8 copies of $x^3+x+1$ . Obviously out of 8 of these terms we can chose it by ether taking 5 $x^3$ terms 1 $x$ term and 2 one terms, or by taking 4 $x^3$ terms and then 4 $x$ terms. In the first case for all the 5 $x^3$ terms we chose we have three ways of combining them totaling in ${8 \choose 5} \times 3$ . In the second case for any 4 $x^3$ terms we chose there is only one way of combining the 4 $x$ term thus in total there are ${8 \choose 4} \times 1$ . Thus on adding all the way of choosing $x^{16}$ from $(x^3+x+1)^8$ are ${8 \choose 5} \times 3+ {8 \choose 4} \times 1=238$ .

[Latex edits - Calvin]

$(x^3+x+1)^8 =[x^3+(x+1)]^8 = \sum_{j=0}^{8}(x^3)^{8-j}(x+1)^j\binom{8}{j} =$ $\sum_{j=0}^{8}(x^3)^{8-j}\sum_{k=0}^{j}\binom{j}{k}x^k\binom{8}{j}$

Thus, we must have that $3\cdot (8-j) +k = 16, thus 8=3j -k$ . The only accettable solutions with $k \leq j$ are $(j=3,k=1)$ and $(j=4,k=4)$ . Inserting this values in the expression, we get that the coefficient of $x^{16}$ is $\binom{8}{3}\binom{3}{1}+\binom{8}{4} = 238$

(x^3+x+1)^8=(x^3+x)^8+8C 1(x^3+x)^7+8C 2(x^3+x)^6+. . . +8C 7(x^3+x)^1+8C 8. by expanding we get the coefficient of x^16. Hint: Just expand first and third term of this expansion. we will get answer

(x3+x+1)^8=[x^24]+[8(x^21)(x+1)]+[36(x^18)(x+1)^2]+[56(x^15)(x+1)^3)]+[70(x^12)(x+1)^4]+[56(x^9)(x+1)^5]+[36(x^6)(x+1)^6)]+[8(x^3)(x+1)^7]+[(x+1)^8]

we have to pick only x^16 terms from the above binomial expansion

56 3(x^16) +70(x^16) 56 3 + 70=168+70=238

Solution 1: The coefficient of $x^ {16}$ arises from $(x^3)^5 \times (x)^1 \times (1)^2$ and from $(x^3)^4 \times (x)^4 \times (1)^0$ . There are no other ways to get $x^{16}$ . Clearly if we have more than 5 $x^3$ terms then the smallest possible exponent is $6\times 3 = 18$ and if we have less than 4 $x^3$ terms then the largest possible exponent is $3\times 3 + 5 = 14$ . For the first case there are ${8 \choose 5}$ ways to pick the $x^3$ term, then there are ${3 \choose 1}$ ways to pick the $x$ term and then there are ${2 \choose 2}$ ways to pick the $1$ term. Thus by the rule of product the coefficient in this case is ${8 \choose 5}\cdot {3 \choose 1} \cdot {2 \choose 2} = 168$ . Similarly, for the second case there are ${8 \choose 4}$ ways to pick the $x^3$ term and then ${4 \choose 4}$ ways to pick the $x$ term. Thus by the rule of product the coefficient in this case is ${8 \choose 4}\cdot {4 \choose 4} = 70$ . Hence the coefficient of $x^{16}$ is $168 + 70 = 238$ .

Solution 2: Applying the binomial theorem, we have $(x^3 + (x + 1))^8 = \sum_{i=0}^{8} {8 \choose i}\times (x^3)^i \times (x+1)^{8-i}$ . If $i \leq 3$ the maximum exponent we can achieve is $3\times 3 + 5 = 14$ . If $i \geq 6$ the minimum exponent we can achieve is $6\times 3 = 18$ . Thus $i = 4$ or $i = 5$ . When $i = 4$ : ${8 \choose 4}\times (x^3)^4 \times (x+1)^4$ and we need the $x^4$ term in $(x+1)^4$ to get $x^{16}$ , and this has a coefficient of $1$ . When $i = 5$ : ${8 \choose 5}\times (x^3)^5 \times (x+1)^3$ , we need the $x$ term in $(x+1)^3$ to get $x^{16}$ , and this has a coefficient of $3$ . Thus, the coefficient of $x^{16}$ is $3\times {8 \choose 5} + {8 \choose 4} = 3(56) + 70 = 238$ .

Note: Both solutions represent the same calculation, though in different terminology. We can also apply the Multinomial Theorem directly, if we knew the formulas.

In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power (x + y)n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. When an exponent is zero, the corresponding power is usually omitted from the term.