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..
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
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
No problem! I can help you with that.
The binomial formula allows us to expand expressions of the form (a + b)^n, where a and b are numbers and n is a positive integer. The formula states that:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
In this case, our expression is (y + 2s)^25, where a = y, b = 2s, and n = 25.
To find the coefficient of the y^24 * s term, we are looking for the term in the expansion of (y + 2s)^25 that contains (y^24) * (2s^1).
The coefficient of this term can be determined using the binomial coefficient, which is given by the combination formula:
C(n, k) = n! / (k! * (n-k)!)
Where n is the total number of terms (25 in this case) and k is the power of the variable we are interested in (1 for the power of s and 24 for the power of y in our case).
So, the coefficient of the y^24 * s term can be calculated as:
C(25, 1) = 25! / (1! * (25-1)!) = 25
Therefore, the coefficient of the y^24 * s term in the expansion of (y + 2s)^25 is 25.