SOLUTION: The first three terms in the binomial expansion of (1+ kx)n are 1− 21x +189x2. Find the value of n and the value of k. Please show all the steps etc... I know the answe

Algebra.Com
Question 367543: The first three terms in the binomial expansion of (1+ kx)n are 1− 21x +189x2.
Find the value of n and the value of k.
Please show all the steps etc... I know the answer, but I don't understand the solution at all... thanks

Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
The first three terms of the expression are . Hence
nk = -21, and
, after equating coefficients.
The foregoing equation is the same as . Hence
by substitution,
, or ,
or 21k = -63, or k = -3. From nk = -21, n = 7.

RELATED QUESTIONS

The binomial (1+kx)^n where n>3. The coefficient of x^2 and x^3 are equal. Find k in... (answered by Edwin McCravy)
If the coefficient of 2nd,3rd and 4th terms in the expansion of (1+x)^n are in A.P, then... (answered by ikleyn)
34. The first three terms in the expansion of (1+ay)^n are 1, 12y, and 68y^2 Evaluate (answered by MathLover1)
The first four terms in the expansion of {{{(1+px)^n}}}, where n > 0, are {{{ 1 + q*x + (answered by solver91311)
The first three terms in the expansion of (1+ay)^n are 1,12y, and 68y^2 Evaluate a and (answered by vleith,adamchapman)
Hello I have 3 questions that are all related and I would greatly appreciate it if... (answered by fractalier)
In the expansion of (2x+1)^n, the coefficient of the x^2 term is 40n , where n is a... (answered by ikleyn)
if the 21st and 22nd terms in the expansion of (1+x)^44 are equal, find the value of... (answered by lynnlo)
These are the questions I'm having problems with: 24.b) (n+2)! / n! = 56 25. solve (answered by nabla)