Question

The coefficient of $x^7$ and $x^8$ in the expansion of $(3+ax{)}^{15}$ are equal they the value of $a$, is $(a\ne 0)$.

• A. $1$
• B. $2$
• C. $3$
• D. None of theseNone of these

Answer 1

Answer: (C)

$3$

The binomial expansion is given by the formula

$(a+b{)}^n={\sum }_{k=0}^n{}^n{C}_k{a}^{n-k}{b}^k$

${(ax+3)}^{15}={\sum }_{k=0}^{15}{}^{15}{C}_k{\left(ax\right)}^{15-k}{\left(3\right)}^k$

According to the question coefficient of $x^7$ and $x^8$ are same so, if we take $k=7$ and $k=8$ we get the coefficent of $x^7$ and $x^8$ ,

$k=7:\frac{15!}{7!(15-7)!}\left(ax{)}^{15-7}\right(3{)}^7=$

$k=8:\frac{15!}{8!(15-8)!}\left(ax{)}^{15-8}\right(3{)}^7$

Coefficient of $x^7$ and $x^8$ are equal so,

$\Rightarrow \frac{15!}{7!8!}{a}^8{3}^7=\frac{15!}{8!7!}{a}^7{3}^8$

$\Rightarrow a^8{3}^7=a^7{3}^8$

$a=3$