The constant term in the expansion of (2x + 1 x ^ 7 + 3 x ^ 2 ) ^ 5...

Any term in the expansion of ${\left(2x + \dfrac{1}{{x}^{7}} + 3{x}^{2}\right)}^{5} is given by</p><p>$\dfrac{5 !}{a ! \cdot b ! \cdot c !}{\left(2x\right)}^{a}{\left({x}^{- 7}\right)}^{b}{\left(3{x}^{2}\right)}^{c}$</p><p> where$a + b + c = 5 . . . \left(i\right)$</p><p>$= {2}^{a}{\left(3\right)}^{c}\left(\dfrac{5 !}{a ! \cdot b ! \cdot c !}\right){x}^{a}{x}^{- 7b}{x}^{2c}$</p><p>$= {2}^{a}{\left(3\right)}^{c}\left(\dfrac{5 !}{a ! \cdot b ! \cdot c !}\right){x}^{a - 7b + 2c}$</p><p>Now for term to be independent of$x$, we know</p><p>$a - 7b + 2c = 0 . . . \left(ii\right)$</p><p>Solving \left(i\right) \text{\text{ and } }$\left(ii\right), we get

$b =$\dfrac{c + 5}{8}

Also, $a, b, c \in \left\{1, 2, 3, 4, 5\right\}$

So, $a = 1, b = 1, c = 3$

Therefore, the term independent of $x$ will be

${2}^{1}{\left(3\right)}^{3} \cdot \dfrac{5 !}{1 ! \cdot 1 ! \cdot 3 !} = 1080$