1. ## Binomial Theorem

Given that $(1+y)^8=1+8y+28y^2+56y^3+...$

In the expansion of $(1+x+kx^2)^8$ in ascending powers of $x$, the coefficient of $y^3$ is zero. Find the value of the constant $k$.

2. Originally Posted by Punch
Given that $(1+y)^8=1+8y+28y^2+56y^3+...$

In the expansion of $(1+x+kx^2)^8$ in ascending powers of $x$, the coefficient of $y^3$ is zero. Find the value of the constant $k$.
Put $y = x+kx^2$ in the first identity we have $\displaystyle (1+x+kx^2)^8=1+8(x+kx^2)+28(x+kx^2)^2+56(x+kx^2)^3 +...$
To have the coefficient of $y^3$ be zero, we must have $56(x+kx^2)^3 = 0 \Rightarrow x+kx^2 = 0 \Rightarrow x(1+kx) = 0 \Rightarrow k = -\frac{1}{x}.$

3. However, the answer is k=-1! and from the second sentence of the question, the expression doesn't contain y, where is there a coeff of y?

4. Originally Posted by Punch
However, the answer is k=-1! and from the second sentence of the question, the expression doesn't contain y, where is there a coeff of y?
I thought they were asking you to consider $y$ as $x+kx^2$ in the expansion $\sum_{r=0}^{8}\binom{8}{r}{y}^r$ to find $k$ given that $\binom{8}{3}(x+kx^2)^3 = 0$,
but it seems not. It's more likely that by the coefficient of $y^3$ they rather meant the coefficient of $x^3$, which gives k as indeed $-1$.

5. This question is weird. Surprisingly, it is a question from past years o levels paper!!! is there a problem with this question then?