Question

The second, third and fourth terms in the expansion (a+b)^n are 12 60 and 160 respectively. Calculate the values of a, b and n

Answer 1

(a+b)^n = a^n + n a^(n-1) b + n(n-1)/2 a^(n-2) b^2 + n(n-1)(n-2)/6 a^(n-3) b^3
so we have
n a^(n-1) b = 12
n(n-1)/2 a^(n-2) b^2 = 60
n(n-1)(n-2)/6 a^(n-3) b^3 = 160
solving that, we arrive at
(1+2)^6 = 1 + 6*2 + 15*2^2 + 20*2^3 + ...= 1+12+60+160+...

Answer 2

Looks like another problem too tough for our AI to tackle

(a+b)^n
= a^n + n a^(n-1) b + n(n-1)/2! a^(n-2) b^2 + n(n-1)(n-2)/6 * a^(n-3) b^3 + ...

n a^(n-1) b = 12
n(n-1)/2! a^(n-2) b^2 = 60
divide the 2nd by the 3rd
simplifies to : 10a = b(n-1)

divide the 3rd by the 4th
simplifies to 9a = 8b(n-2)
dividing [10a = b(n-1)] by [9a = 9b(n-2) ]
gives a value of n = 11

up to you to find values of a and b, hint: they are not nice