Remember Pascal's triangle?
1
11
121 n=2
1331
14641 n=4
1,5,10,10,5,1
1,6,15,20,15,6,1 n=6
your coefficents are 1,30,375
a*something=30
a^2*next coefficent=375
So, your a=6, then n=6 (last row) and verify by a^2*15=375 yes, you are correct.
the Q is :
Given that (1+ax)^n = 1 + 30x +375x^2+.... find the values of the constants a and n .
I know that :
(i am writing (n 1) as the coefficient of n)
(n 1) *ax = 30x
(n 2) *(ax)^2 = 375x^2
which can be written as
(n 1)* a = 30
(n 2)*a^2 = 375
I solved this only by intuition and got n = 6 and a = 5
but does anyone know how to calculate it otherwise?
2 answers
Or
The first 3 terms of (1 + ax)^n
= 1 + n(ax) + n(n-1)/2(a^2x^2) + ...
so nax = 30x
na = 30
a = 30/n
and
n(n-1)/2(a^2x^2) = 375x^2
n(n-1)(900/n^2) = 750
900(n-1) - 750n
n = 5
then a = 30/5 = 6
The first 3 terms of (1 + ax)^n
= 1 + n(ax) + n(n-1)/2(a^2x^2) + ...
so nax = 30x
na = 30
a = 30/n
and
n(n-1)/2(a^2x^2) = 375x^2
n(n-1)(900/n^2) = 750
900(n-1) - 750n
n = 5
then a = 30/5 = 6