Question

If the coefficents of $x^3$ and $x^4$ in the expansion of $(1+ax+b{x}^2)(1-2x{)}^{18}$ in powers of $x$ are both zero, then $(a,b)$ is equal to:

• A. $(14,\frac{272}{3})$
• B. $(16,\frac{272}{3})$
• C. $(16,\frac{251}{3})$
• D. $(14,\frac{251}{3})$

Answer 1

The correct option is B $(16,\frac{272}{3})$

Consider
$(1+ax+b{x}^2)(1-2x{)}^{18}=(1+ax+b{x}^2\left){[}^{18}{C}_0{-}^{18}{C}_1\right(2x\left){+}^{18}{C}_2\right(2x{)}^2-{}^{18}{C}_3\left(2x{)}^3{+}^{18}{C}_4\right(2x{)}^4-...]\text{Coeff. of}{x}^3={}^{18}{C}_3(-2{)}^3+a.(-2{)}^2{.}^{18}{C}_2+b(-2).{}^{18}{C}_1=0\text{Coeff. of}{x}^3={-}^{18}{C}_3.8+a\times 4^{18}{C}_2-2b\times 18=0=-\frac{18\times 17\times 16}{6}\cdot 8+\frac{4a\times 18\times 17}{2}-36b=0=-51\times 16\times 8+a\times 36\times 17-36b=0=-34\times 16+51a-3b=0=51a-3b=34\times 16=544=51a-3b=544.....(i)$
Only option number (b) satisfies the equation number (i)