Solving Binomial Theorem Qs: If nC0 + nC1 + nC2...+ nCn = 256

[angel_eyez]

i an havign trouble solving this qs

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

all help appreciated

 

[StatusX]

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

 

[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 ?

 

[StatusX]

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

 

[HallsofIvy]

Do you understand that the crucial question here is "What is 1+ 1?"

 

[ekinnike]

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

 

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

 

[HallsofIvy]

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, 2^{5= 32, 26= 64, 27= 128, 28= 256. Well, gosh, I guess n= 8 so that 2n=256!}

 

[angel_eyez]

thnx...i get it now