Binomial Q (1 Viewer)

[azureus88]

need help with 2 questions:

1. By considering [maths]1+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n=(1+x)^n[/maths], show that [maths]1-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...+(-1)^n\frac{1}{n+1}\binom{n}{n}=\frac{1}{n+1}[/maths]

2. Prove that [maths]1+\binom{10}{2}3^2+\binom{10}{4}3^4+\binom{10}{6}3^6+\binom{10}{8}3^8+3^{10}=2^9(2^{10}+1)[/maths]

is it just me or is the first question sorta screwd up.

 

[Timothy.Siu]

just you

 

[azureus88]

lol, so any luck with the question? the first one seems like integration but cant seem to sub in nythin... as for the 2nd one, i got up to (1+3)^10 = 2^20 but dont no what to do with the powers of 1,3,5,7,9

 

[addikaye03]

they both work, really cbf typing it out tho, give it an hour if no one has done it i will

 

[azureus88]

you dont have to give me the full solution. just explain how you got there. i think i'll manage.

 

[Timothy.Siu]

i got the first one,
u integrate it and sub in x=-1

 

[azureus88]

i got the first one,
u integrate it and sub in x=-1

u sure? cause when you integrate RHS, you get (1+x)^(n+1)/(n+1), sub in x=-1 and it'll equal 0.

 

[lolokay]

for the first one, when you integrated did you remember the constant of integration?

 

[Timothy.Siu]

u sure? cause when you integrate RHS, you get (1+x)^(n+1)/(n+1), sub in x=-1 and it'll equal 0.

did u remember the constant?

 

[Timothy.Siu]

lol i got beaten, i didn't copy...

 

[lolokay]

for the second, you just have to figure out a way that you could get those odd powers to somehow cancel out. (looking at the RHS may give you an idea of something to try)

 

[azureus88]

ahh farout... stupid constant. nyway, thanks everyone

 

[addikaye03]

need help with 2 questions:

1. By considering [maths]1+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n=(1+x)^n[/maths], show that [maths]1-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...+(-1)^n\frac{1}{n+1}\binom{n}{n}=\frac{1}{n+1}[/maths]

2. Prove that [maths]1+\binom{10}{2}3^2+\binom{10}{4}3^4+\binom{10}{6}3^6+\binom{10}{8}3^8+3^{10}=2^9(2^{10}+1)[/maths]

is it just me or is the first question sorta screwd up.

(1+x)^n=nCo+nC1(x)+nC2(x^2)+nC3(x^3)+...+nCn(x^n)

integrate [(1+x)^n+1]/n+1=nCo(x)+(1/2)nC1(x^2)+(1/3)nC2(x^3)+(1/4)nC3(x^4)+...+(1/n+1)nCn(x^n)+C

Sub in x=-1, then rearrange.

P.S: ill do the rest if u dont get it from there. Basically [(1+x)^n+1/]/n+1 becomes 0 then u rearrange etc.

 

[azureus88]

yeh i got them both now. but thanks nyway

 

[addikaye03]

yeh i got them both now. but thanks nyway

haha yer, i wish i refreshed the page before i starting helping lol i left the page open from since i wrote my first comment lol

 

[azureus88]

haha yer, i wish i refreshed the page before i starting helping lol i left the page open from since i wrote my first comment lol

haha, at least you didnt end up having to do the 2nd one.

 

[addikaye03]

haha, at least you didnt end up having to do the 2nd one.

haha yer, i was gona type out both solutions aswell, with step by step explanation lol thank god i refreshed