Q58 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

The coefficient of $x^5$ in $\left(3+x+x^2\right)^6$ is

  • A
    $18$
  • B
    $540$
  • C
    $1620$
  • Correct answer
    D
    $2178$

Solution

Since, the general term in the expansion $\left(3+x+x^2\right)^6$ $=\frac{6!}{p!q!r!} 3^p \cdot x^q \cdot\left(x^2\right)^r$, where $p+q+r=6$ $=\frac{6!.3^p \cdot x^{q+2 r}}{p!q!r!}$, where, $p+q+r=6$ Since, $q+2 r=5 \Rightarrow q=5-2 r$ and, $p+5-2 r+r=6 \Rightarrow p=1+r$ If $r=0 \Rightarrow p=1, q=5$. If $r=1 \Rightarrow p=2, q=3$ If $r=2 \Rightarrow p=3, q=1$. So, co-efficient of $x^5$ $=\frac{6!3^1}{1!.5!\cdot .0!}+\frac{6!3^2}{2!.3!\cdot 1!}+\frac{6!.3^3}{3!\cdot 1!\cdot 2!}=18+540+1620=2178$

Hence, the correct answer is option: 'D'.

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