2
$\begingroup$

I'm new to math and trying to learn about the Binomial Theorem, by following this tutorial. I got stuck trying to read the Induction Proof.

They give an example of using the Sum notation:

$$ (x + y)^1 = {1 \choose 0} x^{1-0} y^0 + {1 \choose 1} x^{1-1} y^1 = \sum^{1}_{k=0} {n \choose k} x^{n-k}y^k.$$

If I simplify the terms in the middle part, I get: $ x^{1-0} y^0 + x^{1-1} y^1 = x + y $ , which is what I expected from $ (x+y)^1 $ .

However if I try to extend that to $ (x+y)^2 $ :

$$ (x + y)^2 = {2 \choose 0} x^{2-0} y^0 + {2 \choose 1} x^{2-1} y^1 + {2 \choose 2} x^{2-2} y^2. $$

And then simplify the terms, I get: $ x^{2-0} y^0 + x^{2-1} y^1 + x^{2-2} y^2 = x^2 + xy + y^2 $ . But there is only one $xy$ whereas I expected to get $2xy$ . . . as in $ (x + y)^2 = x^2 + 2xy + y^2 $ .

Why is there only $xy$ instead of $2xy$ , am I missing something here ?

$\endgroup$
3
  • 2
    $\begingroup$ You are forgetting to multiply the binomial coefficients. $\endgroup$
    – Sri-Amirthan Theivendran
    Commented Jul 23, 2016 at 0:00
  • $\begingroup$ ${2 \choose 1}=2$ $\endgroup$
    – richarddedekind
    Commented Jul 23, 2016 at 0:13
  • $\begingroup$ Thanks for that @FoobazJohn . I thought I could just ignore those binomial coefficients, that they were just part of the Summation notation. Now I know what they are :-) $\endgroup$
    – user355720
    Commented Jul 23, 2016 at 0:21

1 Answer 1

Reset to default
1
$\begingroup$

The coefficients given by the formula are correct; for instance, ${2}\choose{1}$$=2$, so it gives you $x^2+2xy+y^2$. In general, ${n}\choose{k} $$=$ $$\frac{n!}{k!(n-k)!}$$ e.g. ${2}\choose{1}$$=$ $$\frac{2!}{1!\cdot1!} = \frac{2}{1} = 2$$

$\endgroup$
1
  • $\begingroup$ Thanks for explaining that, it makes sense now. $\endgroup$
    – user355720
    Commented Jul 23, 2016 at 0:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .