If the coefficients of x^2 and x^3 are both zero, in the expansion of the expression (1+ax…

Question

If the coefficients of $x^2$ and $x^3$ are both zero, in the expansion of the expression $(1+ax+bx^2)(1-3x)^{15}$ in powers of $x$, then the ordered pair $(a,b)$ is equal to:

Accepted Answer

Coefficient of $x^2$: $\binom{15}{2}(-3)^2+a\binom{15}{1}(-3)+b\binom{15}{0}=0\Rightarrow945-45a+b=0\quad\text{...(i)}$

Coefficient of $x^3$: $\binom{15}{3}(-3)^3+a\binom{15}{2}(-3)^2+b\binom{15}{1}(-3)=0$

$$15\times3\left[-3\times7\times13+21a-3b\right]=0\quad\Rightarrow\quad21a-b=273\quad\text{...(ii)}$$

Solving (i),(ii): $a=28,\ b=315$

Solutions · 1