Sequence and Series (2) Flashcards
First three terms are (k+4), k and (2k-15), show that k^2 - 7k - 60 = 0
k/(k+4) = (2k-15)/k
k^2 = 2k^2 + 8k - 15k - 60
k^2 - 7k - 60 = 0
k/(k+4) = (2k-15)/k
k^2 = 2k^2 + 8k - 15k - 60
k^2 - 7k - 60 = 0
First three terms are (k+4), k and (2k-15), show that k^2 - 7k - 60 = 0
20C7
20!/(7!13!)
Another way of writing n! in nCr
n(n-1)(n-2)(n-3)!
n(n-1)(n-2)(n-3)!
Another way of writing n! in nCr
In the binomial expansion of (1+x)^n, where n>=4, the coefficient of x^5 is two times greater than the coefficient of x^4.
nC5 = 2 * nC4
In the binomial expansion of (1+5x)^n, where n>=3, the coefficient of x^3 is the difference between 32 times the coefficient of x^2 and 95 times the coefficient of x
nC3 * 5^3 = (32 * nC2 * 5^2) - (95 * nC2 * 5^1)
Find the first five terms in the binomial expansion of (4+3x)^6 in ascending powers of x
(6C0)(4^6)(3x^0) + (6C1)(4^5)(3x) + (6C2)(4^4)(3x^2) + (6C3)(4^3)(3x^3) + (6C4)(4^2)(3x^4)
The coefficient of the x^3 term in the binomial expansion of (1+px)^7 is 280. Find value of p
(7C3)(px)^3
35p^3x^3 = 280x^3
Then solve
Find the coefficient of x^9 in the expansion (4-2x)^11?
(11C9)(4^2)(-2x^9)
Get (4+3x)^(1/2) in the form (1+x)^n
(4^(1/2) (1+(3/4)x)^(1/2))
2(1+(3/4)x)^1/2
Find the binomial expansion of (1+x)^-2 up to and including the term x^3
1 + (-2)x + ((-2)(-3))/12 + ((-2)(-3)(-4))/123
The first three terms of (1+6x+12x^2+8x^3)(1+2x+3x^2+4x^3+…)
1(1+2x+3x^2) + 6x(1+2x) + 12x^2(1)
Get the percentage error
|(real value - estimate)/real value|
|(real value - estimate)/real value|
Get the percentage error