Question

What is the value of the 3rd term of the expansion (3y-2)^7?A. 35a^4b^3B. -10206y^5C. 21a^5b^2D. 20412y^5

Answer 1

Solution

Using the Binomial Theorem

$(A+B)=\sum ^n_{k\mathop=0}{{{{n\choose k}}}}A^{n-k}B^k$

The r-th term of a binomial expansion is given as;

$T_r={{{{{{{n\choose r-1}}}}}}}A^{n+1-r}B^{r-1}$

so when r = 3

$\begin{gathered} T_3={{{{{{{n\choose 2}}}}}}}A^{n-2}B^2 \\ \text{ If }n=7,A=3y\text{ and }B=-2 \\ \Rightarrow T_3={{{{{{{{7\choose 2}}}}}}}}\times(3y)^{7-2}(-2)^2 \\ \Rightarrow T_3=21\times(3y)^5\times4 \\ \Rightarrow T_3=21\times3^5\times y^5\times4 \\ \Rightarrow T_3=20412y^5 \end{gathered}$

Hence the correct option is D