(1+ax)^n has the first three terms of 1+21x+189x^2 what is a and n, how would i go about doing this?

Answer 1

using binomiual theorem
coefficient of x = na = 21 ... 1

coefficeint of x^2 = n(n-1)/2 a^2 = 189

or n(n-1) a^2 = 21 * 18... 2

solve 1 and 2 for n and 1

square 1 and divide by 2

n/(n-1) = 21/18 = 7/6

or n = 7
from 1 a = 3

Answer 2

1+21x+189x^2......(1)
use binomial expansion;
(1+ax)^n
=1+n(ax)+n(n-1)(ax)^2/2 +.....
compare coefficients with (1),
a*n=21>>n=21/a.......(2)
.n(n-1)a^2/2=189
(n^2-n)a^2=378.......(3)
substitute n=21/a into (3),
(442/a^2-21/a)a^2=378
441-21a=378
21a=63>>>a=3
from (1), n=7
hence,the expansion
1+21x+189x^2+.....+
is the first three terms of
(1+3x)^7

i hope that this helps

Answer 3

the t term of (1+ax)^n is given by n!/t!(n-t)! a^t
so substituting t=1. 21 = n!/(n-1)! * a = an
substituting t=2. 189 = n!/2 (n-2)! * a^2 = n(n-1)/2 *a^2

a^2 * n^2 - a * an = 2 * 189 = 378
a = ( 21 ^ 2 - 2 * 189)/ 21 = 3; n = 7

formula is (1 + 3x) ^ 7

Answer 4

Use binomial theorem

=>let y=(1+ax)^n
=>y=1 + nax + [n(n-1)/2!](ax)^2 + ...............+(ax)^n

=>nax=21x
=>na=21.....................(1)

and, n(n-1)/2(ax)^2 = 189x^2
=>[n(n-1)/2]a^2 = 189
=>n(n-1)a^2=378
a=21/n...........................from (1)

=>n(n-1)441/n^2 = 378
=>(n-1)/n=378/441
=>441n - 441=378n
=>63n=441
=>n=7

and a=21/7=3

hence (3,7)

Answer 5

(1+ax)^n = 1 +C(1,n)ax +C(2,n)(ax)^2
So we know
n*a=21
and n(n-1)/2 *a^2 =189
so a=21/n and (n-1)*21^2 =378n
63n= 441 so n=7 and a=3