1. ## binomial

(a) find the 3 terms in ascending powers of x of binomial expansion of
(1+px)^9

where p is a constant.

these first 3 terms are 1,36x and qx^2, where q is a constant.

(b)find the value of p and value of q.

2. Originally Posted by outlaw
(a) find the 3 terms in ascending powers of x of binomial expansion of
(1+px)^9

where p is a constant.

these first 3 terms are 1,36x and qx^2, where q is a constant.

(b)find the value of p and value of q.
by the binomial expansion theorem, $(1 + px)^9 = \sum_{k = 0}^{9} { 9 \choose k} 1^{9 - k}(px)^k$

use that to find the first three terms, then you can equate them to the terms you were given to solve for the unknowns

3. by the binomial expansion theorem,

use that to find the first three terms, then you can equate them to the terms you were given to solve for the unknowns

i have found the 3 terms, they are 1+9px+36(px)^2

dont know how to equate....

when i equate there are too many unknowns

4. Originally Posted by outlaw
(a) find the 3 terms in ascending powers of x of binomial expansion of
(1+px)^9

where p is a constant.

these first 3 terms are 1,36x and qx^2, where q is a constant.

(b)find the value of p and value of q.
Originally Posted by outlaw
by the binomial expansion theorem,

use that to find the first three terms, then you can equate them to the terms you were given to solve for the unknowns

i have found the 3 terms, they are 1+9px+36(px)^2

dont know how to equate....

when i equate there are too many unknowns
So
$1 = 1$ <-- Easy one. I like solving these!

$9p = 36$

$36p^2 = q$

So from the second equation we get p = 4, and inserting that into the third equation we get q = 576.

-Dan