Testing the BSL Knowledge on binomial coefficients.

[Chief2ndzOnly!]

I got time so I'd participate in this cumbersome thread.

First, since the exponent of the binomial is 5 then we list the powers of a in descending order (assume the number on the right as exponent).

a5 a4 a3 a2 a

Then insert starting on the right of the 2nd term ascending powers of b and add an extra term b5.

a5 a4b a3b2 a2b3 ab4 b5

Now for the coefficients, we use the 6th row of the pascal triangle.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

and insert them in our previous expression except for the 1's.

a5 5a4b 10a3b2 10a2b3 10ab4 b5

Now insert + if it's the sum of the binomial and alternating + - if the difference of a binomial. Hence,

(a+b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 10ab4 + b5

Ther's another mechanical technique which I'm willing to share upon request.

Sweet Jesus Kad. I know EXACTLY where I'm goin if I EVER need help with math.

[Barry Halls]

Sweet Jesus Kad. I know EXACTLY where I'm goin if I EVER need help with math.

lucky for you, there's wolframalpha.com

[Chief2ndzOnly!]

lucky for you, there's wolframalpha.com

Great. With the help of the internet, I can become a mathmatical genius!!!!!!!!!! Thanks Harry.

[kadyo's]

Sweet Jesus Kad. I know EXACTLY where I'm goin if I EVER need help with math.

Anytime, chef, anytime.

[kadyo's]

lucky for you, there's wolframalpha.com

I tried looking for an answer to this question in that site and they were not able to give me a general term which is understandable but I was expecting them to talk to me about bernoulli's numbers at least.

[baCCaT]

I got time so I'd participate in this cumbersome thread.

First, since the exponent of the binomial is 5 then we list the powers of a in descending order (assume the number on the right as exponent).

a5 a4 a3 a2 a

Then insert starting on the right of the 2nd term ascending powers of b and add an extra term b5.

a5 a4b a3b2 a2b3 ab4 b5

Now for the coefficients, we use the 6th row of the pascal triangle.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

and insert them in our previous expression except for the 1's.

a5 5a4b 10a3b2 10a2b3 10ab4 b5

Now insert + if it's the sum of the binomial and alternating + - if the difference of a binomial. Hence,

(a+b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 10ab4 + b5

Ther's another mechanical technique which I'm willing to share upon request.

difficulty level : Asian

nice one manong kadz

[crookshanks]

a simple equation. no degree is needed to solve this.

x=√(2+√(2+√(2+………∞

[Spartacus Sully]

a simple equation. no degree is needed to solve this.

x=√(2+√(2+√(2+………∞

is the answer ∞?

No NO wait, its 2

[crookshanks]

is the answer ∞?

No NO wait, its 2

and......................

[Spartacus Sully]

and......................

and?

and what?

its not quite 2 but its infinatly close to 2 in such a way that reguardless of how close to 2 a number is that you give me i can use the equation to create a number closer to 2.

thus in math you can think of it as being 2, but really its just infanity close to 2.