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

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}}}}}}}}*(3y)^(7-2)(-2)^2 \\ \Rightarrow T_3=21*(3y)^5*4 \\ \Rightarrow T_3=21*3^5* y^5*4 \\ \Rightarrow T_3=20412y^5 \end{gathered}`

Hence the correct option is D