# C2 Binomial Expansion help plz :)Watch

Thread starter 12 years ago
#1
:O Help plz!

Given that

(2 - x)13 = A + Bx + Cx2 + ... ,

find the values of the integers A, B and C.

N.B. it shouldnt be an = sign but 3 lines meaning an identity or whatever :P
0
12 years ago
#2
A = 2^13 = 8192
B = 13C1 * 2^12 * (-x) = -13*4096 = -53248
C = 13C2 * 2^11 * (-x)^2 = 78*2048 = 159744

quick check in calculator: x=0.01
0
Thread starter 12 years ago
#3
thanks dude, i got another one if you can help :P
0
Thread starter 12 years ago
#4
Can anyone help with this one?

Expand (1 - 2x)10 in ascending powers of x up to and including the term in x3, simplifying each coefficient in the expansion.
Use your expansion to find an approximation to (0.98)10, stating clearly the substitution which you have used for x.
0
12 years ago
#5
(1-2x)^10

= 1^10 + 10C1 * 1^9 * (-2x) + 10C2 *1^8 * (-2x)^2 + 10C3 *1^7 * (-2x)^3
= 1 - 20x + 180x^2 -960x^3

to approximate 0.98^10
let x = 0.01
0.98^10 = 1-20(0.1) + 180(0.01)^2 - 960(0.01)^3
0.98^10 = 0.81704
1
Thread starter 12 years ago
#6
The coefficient of x2 in the binomial expansion of (1 + x/2) n, where n is a positive integer, is 7.

Find the value of n.
Using the value of n found in (a), find the coefficient of x4.

Last question i got for now

btw - left you positive rep thanks again for the help
0
12 years ago
#7
(1+0.5x)^n

coeff. of x^2 = 7 = 1/2* n(n-1)* 0.5^2
56 = n^2-n
n = 8 as n is positive

coeff of x^4:
8C4 * 1^4 * (x/2)^4
= 35/4
0
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