In a nutshell:
the general term of the expansion of (a+b)^nis
termr+1 = C(n,r) a^(n-r) b^r
so for (y+2s)^25
termr+1 = C(25,r) y^(25-r) (2s)^r
so you want the term containing y^24 s
that is 25-r = 24
r = 1
that is we want term2 which is
C(25,1) y^24 (2s)^1
= 25 y^24 (2s) = 50 y^24 s
thus the coefficient of your required term is 50
Use the binomial formula to find the coefficient of the y^24 s
term in the expansion of (y+2s)^25
(My teacher told me this wasn't going to be on the final but it was and he never taught us it!!!!!!) I cant afford to guess again..
2 answers
or
since the term y^24 is "close" to 25, why not just find the first few terms
(y + 2s)^25
= y^25 + 25 y^24 (2s ) + 25(24)/2! y^23 (2s)^2 + ...
= y^25 + 50 y^24 s + 1200 y^23 s^2 + ...
ahhh, we have it
check:
https://www.wolframalpha.com/input/?i=expand+(y+%2B+2s)%5E25
look at the second-last term
since the term y^24 is "close" to 25, why not just find the first few terms
(y + 2s)^25
= y^25 + 25 y^24 (2s ) + 25(24)/2! y^23 (2s)^2 + ...
= y^25 + 50 y^24 s + 1200 y^23 s^2 + ...
ahhh, we have it
check:
https://www.wolframalpha.com/input/?i=expand+(y+%2B+2s)%5E25
look at the second-last term