JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 16)
The constant term in the expansion of $${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$$ is ___________.
Answer
1080
Explanation
Constant term in the expansion of
$$ \begin{aligned} & \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5} \end{aligned} $$
Term independent of $x=$ coefficient of $x^{35}$ in
$$ \begin{aligned} & ^{5} C_{4}\left(x^{8}(3 x+2)\right)^{4} \\\\ = & { }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4} \\\\ = & { }^{5} C_{4} \times{ }^{4} C_{3}(2)^{1}(3)^{3} \\\\ = & 5 \times 4 \times 2 \times 27 \\\\ = & 1080 \end{aligned} $$
$$ \begin{aligned} & \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5} \end{aligned} $$
Term independent of $x=$ coefficient of $x^{35}$ in
$$ \begin{aligned} & ^{5} C_{4}\left(x^{8}(3 x+2)\right)^{4} \\\\ = & { }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4} \\\\ = & { }^{5} C_{4} \times{ }^{4} C_{3}(2)^{1}(3)^{3} \\\\ = & 5 \times 4 \times 2 \times 27 \\\\ = & 1080 \end{aligned} $$
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