# Binomial theorum

1. Jun 16, 2006

### angel_eyez

i an havign trouble solving this qs

if nC0 + nC1 + nC2 +...+ nCn = 256 find the value of n

all help appreciated

2. Jun 16, 2006

### StatusX

You know the binomial theorem, so what values of x and y have the expansion of (x+y)^n equal to that sum?

3. Jun 16, 2006

### angel_eyez

i don't get it, i know that teh values of x and y shoudl be one, but if it is equal to 256 how can i put that ?

4. Jun 16, 2006

### StatusX

Right, so what does n have to be for (1+1)^n to equal 256?

5. Jun 17, 2006

### HallsofIvy

Staff Emeritus
Do you understand that the crucial question here is "What is 1+ 1?":rofl:

6. Jun 18, 2006

### ekinnike

its a series question . i forgot how to do it..sorry.

7. Jun 18, 2006

### ekinnike

cool i solved it lol. took less than 5 mins tried and error on calc.. there is a proper way to solve it... well n=8 i work it out by put numbers into n@_@ yeh 8 is correct value.

omfg im sorry guys it oready been solve... by (1+1)^n=256 <--- how did dat work out@_@ well i did tried my best@_@

Last edited: Jun 18, 2006
8. Jun 18, 2006

### HallsofIvy

Staff Emeritus
Excuse me? 'Trial and error'? The whole question was "for what n does n does 2n= 256. How long does that take to calculate?
22= 4, 23= 8, 24= 16, 25= 32, 26= 64, 27= 128, 28= 256. Well, gosh, I guess n= 8 so that 2n=256!

9. Jun 18, 2006

### angel_eyez

thnx...i get it now