How do you make a proof using PMI for
(n) + (n) + (n)+...+(n) = 2^n
0 1 2 n
using the binomial theorom?
I don't know how to type the parentheses around the whole n and 0, n and 1, n and 2, and n and n expressions. In other words, there should be one large parenthesis around each of the terms above.
If you throw a coin one time,the possiblities of outcome are 2= 2^1 Head or Tail
If you throw a coin 2 times, possiblities of outcome are: hh, th,ht,tt=4 =2^2 possiblities.
Similarly if the coin is thown 3 times , possblities are:hhh,hht,hth,thh,htt,tht,tth,ttt = 2^3.
Thus each trial has 2 possible outcomes.
If the coin is thrown n times , the number possible outcomes are =2^n.
Let us see the other way these outcomes of n throws:
In n trials, number no heads
No heads and all are tails in ntrials is only one possiblityand symbolically it is nC0.
Number one head comining up of one head = nC1
Number of ways of appearing 2 heads =nC2
Number of ways of appearing 3 heads =nC3
Number of ways of appearing 4 heads =nC4
........... ......... .................... .............
Number of ways of appearing all n heads=nCn.
Adding all the above ways heads from o heads to n heads in n trials exhausts all ways possiblities which is 2^n. Therefore,
nC0+nC1+nC2+nC4+................+nCn = 2^n
In order to prove that nC0+nC2+nC3+.....+nCn=2^n.
We must first learn to derive the binomial series.
If we put x=1 and y=1
(1+1)^n=2^n= nc0+nc1+nc2+...+nCr+...+nCn (shown)
In order to prove that (Cn)0 + (Cn)1 +....+(Cn)n=2^n, you have to use Newton's binomial development:
If we'll make "n" term(in paranthesis) equal to 1, we'll have:
But the digit 1, raised at any exponent, will remain equal to 1.
(1+1)^n=2^n=(Cn)0 + (Cn)1 + (Cn)2+...+(Cn)k+...+(Cn)n