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

Much thanks in advance! :)

Question

Expand (1 + y)^8 in ascending powers of y, up to y^3.
In the expansion (1 + x + kx)^8 in ascending powers of x, the coefficient of x^3 is zero. Find the constant k.

Much thanks in advance! :)

anonymous · Answer

$(1+y)^8=1+8y+28y^2+56y^3+...$ coefficients come from $\binom{8}{k}$ for $k=0,1,2,3$

anonymous · Answer

so for example the coefficient of the $y^2$ term is $\binom{8}{2}=\frac{8\times 7}{2}=4\times 7=28$

anonymous · Answer

was this really posted ten hours ago?

anonymous · Answer

@satellite73 I'm sorry for replying to a week-old post. D: I haven't had the time to go back to it lately, I'm really really sorry. Thank you so much for your help!