(1+x)^(n+1) = 1^(n+1) + (n+1)1^n x
+ (n-1)(n-2)/2 1^(n-1)x^2 + (n+1)n(n-1)/6 1^(n-2) x^3 +(n+1)n(n-1)(n-2)/24 1^(n-3)x^4
huum
(n+1)n(n-1)(n-2)/24 = 6 k
(1+x)^(n-1) = 1^(n-1) + (n-1)1^(n-2) x + (n-1)(n-2)1^(n-3)/2 * x^2
so
(n-1)(n-2)/2 = k
(n-1)(n-2) = 2 k
so
(n+1)n (2k)/24 = 6 k
n^2 + n =72
n^2 + n - 72 = 0
(n-8)(n+9) = 0
n = 8
etc
in the expansion of (1+x)^(n+1) the coefficient of x^4 is 6k and in the expansion of (1+x)^(n-1) the coefficient of x^2 is k.find the value of k and n, for n>2???
7 answers
what how did you got 72 what happen to the k
(n+1)n(n-1)(n-2)/24 = 6 k
is
(n+1)n [(n-1)(n-2) ]= 72 * 2 k
BUT
[(n-1)(n-2)] = 2 k
so
(n+1) n = 72
is
(n+1)n [(n-1)(n-2) ]= 72 * 2 k
BUT
[(n-1)(n-2)] = 2 k
so
(n+1) n = 72
got it thanks
You are welcome.
screw down sir there is a question on parametrization and explain well better although sir steve has done much explanation but i wanna get more ideal
He said:
Actually, just consider the definition of T.
Take a peek here, and it should be clear, since the arc length is just
∫ ds = ∫ |dr/dt| dt
88888888888888888888888888
In other words the tangent is along the curve so if you integrate the tangent along the curve you get the length.
Actually, just consider the definition of T.
Take a peek here, and it should be clear, since the arc length is just
∫ ds = ∫ |dr/dt| dt
88888888888888888888888888
In other words the tangent is along the curve so if you integrate the tangent along the curve you get the length.