Question

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???

Answer 1

(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

Answer 2

what how did you got 72 what happen to the k

Answer 3

(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

Answer 4

got it thanks

Answer 5

You are welcome.

Answer 6

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

Answer 7

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.