# Binomial Expansion Question Help PleaseWatch

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#1
the first three terms in asscending powers of x of the binomial exapansion (1+bx)^n are 1 + 28x + 336x^2

n is a psotive interger find the value of n and b

good luck

if it is any help i got as far as 1. bx^n = 28x and 2. 672=(n^2-n)b^2
0
13 years ago
#2
(1+bx)n = 1 + nbx + Â˝n(n-1)b2x2 +...
=> nb = 28....................(1)
=> Â˝n(n-1)b2 = 336...(2)
In (1): n=28/b
Sub. in (2):
=> Â˝(28/b)(28/b-1)b2 = 336
Â˝(28)(28-b) = 336
28-b = 24
b = 4
Sub. in (1):
4n = 28
n = 7
0
13 years ago
#3
(Original post by edwin8030)
the first three terms in asscending powers of x of the binomial exapansion (1+bx)^n are 1 + 28x + 336x^2

n is a psotive interger find the value of n and b

good luck

if it is any help i got as far as 1. bx^n = 28x and 2. 672=(n^2-n)b^2
(1+bx)^n=1+nbx+n(n-1)/2.(bx)^2.....
(1+bx)^n=1+(x)(nb)+(x^2)[0.5n(n-1)b^2].....
(1+bx)^n=1+28x+336x^2.....
By equating coefficients: nb=28 and 0.5n(n-1)b^2=336
b^2=(28/n)^2 and b^2=336/0.5n(n-1)
(28/n)^2=336/0.5(n-1).
Solve to find n and hence b.
0
13 years ago
#4
Aww I just wrote out a solution.. and I've been beaten to it
0
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