Question:

Step 1: Write down the general term for the trinomial expansion. For an expansion of the form $(A+B+C)^n$, the general term is given by the multinomial theorem as $\dfrac{n!}{p!q!r!} A^p B^q C^r$, where $p+q+r=n$ and $p,q,r$ are non-negative integers. In this problem, we have $\left(2x+\dfrac{1}{x^7}+3x^2\right)^5$. Let $A=2x$, $B=x^{-7}$, $C=3x^2$, and $n=5$. The general term is: $$T_{p,q,r} = \dfrac{5!}{p!q!r!} (2x)^p \left(\frac{1}{x^7}\right)^q (3x^2)^r$$ Step 2: Simplify the general term and identify the exponent of $x$. Substitute the terms and simplify the powers of $x$: $$T_{p,q,r} = \dfrac{5!}{p!q!r!} (2^p x^p) (x^{-7q}) (3^r x^{2r})$$ $$T_{p,q,r} = \dfrac{5!}{p!q!r!} 2^p 3^r x^{p - 7q + 2r}$$ For the term to be constant, the exponent of $x$ must be zero. Step 3: Set up the equations for the constant term. We need to find non-negative integer values for $p, q, r$ that satisfy two conditions: 1. The sum of the powers must equal $n$: $$p+q+r=5 \quad (*)$$ 2. The exponent of $x$ must be zero: $$p - 7q + 2r = 0 \quad (**)$$ Step 4: Solve the system of equations for $p, q, r$. From equation $(*)$, we can express $p$ as $p = 5-q-r$. Substitute this into equation $(**)$: $$(5-q-r) - 7q + 2r = 0$$ $$5 - 8q + r = 0$$ $$r = 8q - 5$$ Since $r$ must be a non-negative integer ($r \ge 0$), we have $8q - 5 \ge 0$, which implies $8q \ge 5$, so $q \ge \frac{5}{8}$. Since $q$ must be an integer, the smallest possible value for $q$ is $1$. Let's test possible values for $q$: Case 1: If $q=1$ Substitute $q=1$ into $r = 8q - 5$: $r = 8(1) - 5 = 3$ Now, substitute $q=1$ and $r=3$ into $p+q+r=5$: $p+1+3=5$ $p+4=5$ $p=1$ So, $(p,q,r) = (1,1,3)$ is a valid set of non-negative integers. Let's verify with $p - 7q + 2r = 0$: $1 - 7(1) + 2(3) = 1 - 7 + 6 = 0$. This confirms the values. Case 2: If $q=2$ Substitute $q=2$ into $r = 8q - 5$: $r = 8(2) - 5 = 16 - 5 = 11$ Now, check if $p+q+r=5$ can be satisfied: $p+2+11=5$ $p+13=5$ $p = -8$ Since $p$ must be non-negative, this case is not possible. Any higher value of $q$ will also result in $r$ being larger, making $p$ negative, or $p+q+r > 5$ even if $p$ is assumed to be minimum $0$. Thus, $(1,1,3)$ is the only possible set of values for $(p,q,r)$. Step 5: Calculate the constant term. Substitute $p=1, q=1, r=3$ into the general term formula: $$T_{1,1,3} = \dfrac{5!}{1!1!3!} 2^1 3^3$$ $$T_{1,1,3} = \dfrac{5 \times 4 \times 3 \times 2 \times 1}{(1)(1)(3 \times 2 \times 1)} \times 2 \times 27$$ $$T_{1,1,3} = \dfrac{120}{6} \times 2 \times 27$$ $$T_{1,1,3} = 20 \times 2 \times 27$$ $$T_{1,1,3} = 40 \times 27$$ $$T_{1,1,3} = 1080$$ Step 6: State the final answer. The constant term in the expansion of $\left(2x+\dfrac{1}{x^7}+3x^2\right)^5$ is $1080$. The final answer is $\boxed{1080}$.