Given that $\displaystyle (1+y)^8=1+8y+28y^2+56y^3+...$
In the expansion of $\displaystyle (1+x+kx^2)^8$ in ascending powers of $\displaystyle x$, the coefficient of $\displaystyle y^3$ is zero. Find the value of the constant $\displaystyle k$.
Given that $\displaystyle (1+y)^8=1+8y+28y^2+56y^3+...$
In the expansion of $\displaystyle (1+x+kx^2)^8$ in ascending powers of $\displaystyle x$, the coefficient of $\displaystyle y^3$ is zero. Find the value of the constant $\displaystyle k$.
Put $\displaystyle y = x+kx^2$ in the first identity we have $\displaystyle \displaystyle (1+x+kx^2)^8=1+8(x+kx^2)+28(x+kx^2)^2+56(x+kx^2)^3 +...$
To have the coefficient of $\displaystyle y^3$ be zero, we must have $\displaystyle 56(x+kx^2)^3 = 0 \Rightarrow x+kx^2 = 0 \Rightarrow x(1+kx) = 0 \Rightarrow k = -\frac{1}{x}.$
I thought they were asking you to consider $\displaystyle y$ as $\displaystyle x+kx^2 $ in the expansion $\displaystyle \sum_{r=0}^{8}\binom{8}{r}{y}^r$ to find $\displaystyle k$ given that $\displaystyle \binom{8}{3}(x+kx^2)^3 = 0$,
but it seems not. It's more likely that by the coefficient of $\displaystyle y^3$ they rather meant the coefficient of $\displaystyle x^3$, which gives k as indeed $\displaystyle -1$.