Question

One of the terms of the expansion of (x + 2y)^10 is ax^8 y^2. Find the value of a.

Answer 1

\[\binom{10}{2}x^8(2y)^2\] would be the term
\((2y)^2=4y\) and \(\binom{10}{2}=\frac{10\times 9}{2}=5\times 9=45\)
so the coefficient will be \(4\times 45\)

Answer 2

oops i think i meant \((2y)^2=4y^2\) but that doesn't change the coefficient

Answer 3

use the binomial theorem:$$(x+2y)^{10}=\sum_{k=0}^{10}\binom{10}kx^{10-k}(2y)^k=\sum_{k=0}^{10}\binom{10}k2^kx^{10-k}y^k$$

Answer 4

hence of course the coefficient of \(x^{10-k}y^k\) just \(\dbinom{10}k2^k\)... here we have \(x^8y^2\) hence our coefficient is \(\dbinom{10}22^2=4\times\dfrac{10!}{2!8!}=2\times10\times9=180\)

Answer 5

Thank you (=