Fichas
Binomial Expansion
Fichas del conjunto (18)
- 1What is pascals triangle?
- 2What is the link between (a+b)^5 and r=5 of pascals triangle?The integers before the letters are the same and in the same order;
a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5
(1 , 5 , 10 , 10 , 5 , 1) - 3When we expand (a+b)^n using binomial expansion, the powers of a and b must add up to.....n
(e.g. (a+b)^3 = a^3 + a^2b + ab^2 + b^3 ; 3=3 , 2+1=3 , 1+2=3 , 3=3) - 4The top number 1 in pascals triangle is called what?The zero position (r=0)
- 5What is the pattern of a row of pascals triangle? (e.g. r=4)1 4 6 4 1
1 x 4/1 = 4 , 4 x 3/2 = 6 , 6 x 2/3 = 4 , 4 x 1/4 = 1
Therefore the pattern is the numerator starts at the row number (4) and decreases down to 1 , and the denominator starts at 1 and increases up to the row number (4) - 6What is the general formula for the nth row when n is a positive integer?1 , n , n(n-1)/1x2 , n(n-1)(n-2)/1x2x3 , n(n-1)(n-2)(n-3)/1x2x3x4 .....
As:
1 x n/1 = n , n x n-1/2 =n(n-1)/1x2 , n(n-1)/1x2 x n-2/3 =n(n-1)(n-2)/1x2x3 , etc....
(follows same pattern as previous) - 7What does factorial mean? (n!)Add the numbers multiplied up to n. For example 5! = 1x2x3x4x5 = 120 , so 5!=120
- 8Expand (z^2-1)^5r=5 is 1 5 10 10 5 1
So a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 , and a=z^2 and b=-1, so sub in these values and simplify to get 2z^10-5z^8+10z^6-10z^4+5z^2-1 - 9[n][r] or nCr (which are binomial coefficients)= ....n!/r!(n-r)!
- 10n!/r!(n-r)! simplifies to give what?The integer before a coefficient
(if it was a^4b , then [n][r] = [5][1] so 5!/1!(5-1)! = 5!/1!4! , 1x2x3x4x5 / 1x1x2x3x4 =5, so the positive integer = 5 - 11In [n][r] , what do n and r represent?n = original power of expansion , r = the power of the 2nd coefficient
(e.g. a^5 = [5][0] and a^2b^3 = [5][3]) - 12If n=10 and r=6 , what is nCr?210 (as 10C6)
- 13What is the general formula for (a+b)^2 ?[n][0]a^n + [n][1]a^n-1b + [n][2]a^n-2b^2 + ..... [n][r]a^n-rb^r
- 14nC0 = ....1
- 15nC1 = ....n
- 16nC2 = ....n(n-1)/2
- 17To find something such as 9C4 without a calculator, what must we do?Use n!/r!(n-r)! and cancel terms ;
1x2x3x4x5x6x7x8x9 / 1x2x3x4x1x2x3x4x5
Here we can cancel this to get 6x7x9 / 1x3
Furhtermore this cancels to 2x7x9/1 to get 126. - 18If the first 3 terms of (3-x/2)^8 are: 6561 - 8748x + 5103x^2 Estimate the value of 2.995^8Say 3-x/2 = 2.995 , so 3-2.995 = x/2 , so x=0.01
Now sub into expansion:
6561 - 8748(0.01) + 5103(0.01)^2 = 6474.0303