text: # Help proving with the Binomial Theorem
  
  1. Apr 30, 2010
  
  ### steveT
  
  1. The problem statement, all variables and given/known data
  
  (n¦0)-(n¦1)+(n¦2)-. . . ± (n¦n)=0
  
  that reads n choose zero and so on
  2. Relevant equations
  
  Prove this using the binomial theorem
  
  3. The attempt at a solution
  
  I really have no idea where to start. Any help would be greatly appreciated
  
  thanks
  1. The problem statement, all variables and given/known data
  
  2. Relevant equations
  
  3. The attempt at a solution
  
  2. Apr 30, 2010
  
  ### Martin Rattigan
  
  Start by writing down the binomial theorem and seeing how you might get the expression you've typed in out of one side or the other.
  
  3. Apr 30, 2010
  
  ### vela
  
  Staff Emeritus
  What does the binomial theorem tell you? Start there.
  
  4. Apr 30, 2010
  
  ### Martin Rattigan
  
  That as well.
  
  5. Apr 30, 2010
  
  ### steveT
  
  Well I've been staring at this thing for the past hour and I'm not coming up with anything. Am I to be looking at the (x+y)^n side of the binomial theorem or the side with the summation
  
  6. Apr 30, 2010
  
  ### vela
  
  Staff Emeritus
  Well, which side looks more like
  
  $$\begin{pmatrix}n\\0\end{pmatrix}-\begin{pmatrix}n\\1\end{pmatrix}+\cdots\mp\begin{pmatrix}n\\n-1\end{pmatrix}\pm\begin{pmatrix}n\\n\end{pmatrix}$$
  
  7. Apr 30, 2010
  
  ### gauss^2
  
  steveT, if you write vela's sum expression in sigma notation, the result should jump right out at you.
  
  8. Apr 30, 2010
  
  ### vela
  
  Staff Emeritus
  You might also want to post what expression you have for the binomial theorem. There are different ways to write it, some more suggestive than others.
  
  9. Apr 30, 2010
  
  ### steveT
  
  This is the expression I'm using
  
  (x+a)^n=∑_(k=0)^n▒〖(n¦k) x^k a^(n-k) 〗
  
  10. Apr 30, 2010
  
  ### vela
  
  Staff Emeritus
  OK, so what values could you plug in for x and a such that you'd get the alternating sign but otherwise have them disappear?
  
  (It might help you to expand the summation to make the comparison more straightforward.)
  
  11. Apr 30, 2010
  
  ### steveT
  
  x=1 and a=0 ?
  
  12. Apr 30, 2010
  
  ### VeeEight
  
  Try plugging that into your equation for the binomial theorem and see what you get. What accounts for the alternating sign?
  
  13. Apr 30, 2010
  
  ### gauss^2
  
  I know the sigma notation can be a bit sketchy when one first learns it, so let me post this: What does this sum below equal? (according to the Binomial Theorem)
  
  $$\sum_{k=0}^n \binom{n}{k}(-1)^k$$
  
  When dealing with sums the dot-dot-dots leave things a bit ambiguous. When you convert a sum with ... into something explicit using sigma notation, it usually makes things a lot easier.
  
  14. May 1, 2010
  
  ### steveT
  
  So when k is even you get plus and when its odd you get minus which accounts for the alternating sign.
  
  15. May 1, 2010
  
  ### VeeEight
  
  Expand the summation that Gauss^2 posted. What is it equal to in terms of (x+y)n? You are making this much harder than it has to be.
  
  You know that one side of the equation is zero. When is (x+y)n = 0? Use this along with the binomial theorem. Once you have your x and y, plug them into the formula for the binomial theorem to see if you do in fact get your desired alternating sum.
  
  16. May 1, 2010
  
  ### gauss^2
  
  Yes.
  
  17. May 3, 2010
  
  ### steveT
  
  Thanks everyone for your help. I UNDERSTAND
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}