How do you make a proof using PMI for (n) + (n) + (n)+...+(n) = 2^n 0 1 2 n using the binomial theorom?I...

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