1. We need the constant term in $\left(2x+\frac{1}{x^7}+3x^2\right)^5.$

2. Let the terms selected from the 5 factors be:

• $a$ times $2x$
• $b$ times $x^{-7}$
• $c$ times $3x^2$

Then $a+b+c=5.$

3. For the resulting term to be constant, the total power of $x$ must be $0$: $a-7b+2c=0.$

So we solve $a+b+c=5, \qquad a-7b+2c=0$ for non-negative integers $a,b,c$.

4. From $a=5-b-c$, substitute into the second equation: $5-b-c-7b+2c=0$ $5-8b+c=0$ $c=8b-5.$

Then $a=5-b-(8b-5)=10-9b.$

Since $a,c\ge 0$, check possible integer values of $b$:

• If $b=0$, then $c=-5$ impossible.
• If $b=1$, then $c=3$, $a=1$ valid.
• If $b\ge 2$, then $a=10-9b<0$ impossible.

So the only valid choice is $a=1,\quad b=1,\quad c=3.$

5. Hence the constant term comes from choosing:

• one $2x$
• one $\dfrac{1}{x^7}$
• three $3x^2$

Its coefficient is $\frac{5!}{1!1!3!}(2)^1(1)^1(3)^3.$

6. Compute: $\frac{5!}{1!1!3!}=\frac{120}{6}=20,$ $2\cdot 3^3=2\cdot 27=54.$

Therefore, $\text{constant term}=20\cdot 54=1080.$

7. Comparison with stored answer:

• Derived answer: $1080$
• Stored correct answer: $1080$

They match.