IloveManUtd
New member
- Joined
- Jul 27, 2010
- Messages
- 48
In the expansion of (3+4x)[sup:33uo6uz1]n[/sup:33uo6uz1], the coefficients of x[sup:33uo6uz1]4[/sup:33uo6uz1] and x[sup:33uo6uz1]5[/sup:33uo6uz1] are in the ratio 5:16. Find the value of n.
Do I do like this: [[n(n-1)(n-2)(n-3)(n-4)]/(5!)] / [[n(n-1)(n-2)(n-3)] /(4!)] = 16/5 ? If yes then what next? I'm stuck. Thx.
Do I do like this: [[n(n-1)(n-2)(n-3)(n-4)]/(5!)] / [[n(n-1)(n-2)(n-3)] /(4!)] = 16/5 ? If yes then what next? I'm stuck. Thx.