# 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...

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.

### 3 Answers | Add Yours

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.

(x+y)^n=nC0x^ny^0+nC1x^n-1y^1+nC2x^n-2y^2+nC3x^n-3y^3+......+nCrx^n-ry^r+...+nCn-1xy^n-1+nCnx^0y^n

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:

(1+n)^n=(Cn)0*1^n+(Cn)1*1^(n-1)*n^1+(Cn)2*1^(n-2)*n^2+...+(Cn)k*1^(n-k)*n^k+...(Cn)n*1^0*n^n

If we'll make "n" term(in paranthesis) equal to 1, we'll have:

(1+1)^n=2^n=(Cn)0*1^n+(Cn)1*1^(n-1)*1^1+(Cn)2*1^(n-2)*1^2+...+(Cn)k*1^(n-k)*1^k+...(Cn)n*1^0*1^n

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