(nC0) + (nC1) + (nC2) + ...... + (nCn) = 2^n...?

Question

Is there any way to prove this formula other than substituting random numbers?....

Answer 1

from n objects one can choose 0 object in

(nc0) ways
one object in (nc1) ways
n objects (ncn) ways

so number of ways = (nC0) +(nC1) + ... + (nCn)

now in each of the selection an object is there or not there

for each object there are 2 ways

for n objects there are 2^n ways
we have found the 2 ways the result
so both are same

(nC0) +(nC1) + ... + (nCn) = 2^n

Answer 2

Yes man.
you know :
(1+x)^n = (nC0)x^0 + (nC1)x^2 + (nC2)x^2 + ...... + (nCn)x^n.
Yes, now put x = 1.
That's the only way to get the formula. It's actually not a formula, it's a conclusion.If you're not satisfied with that, there's nothing else that will satisfy you.

Answer 3

Draw out pascal's. You'll find each number goes to the next row twice, meaning the total sum doubles.

Then, by induction, you only need to check n=0, which is obviously true.

Answer 4

enhance (a million+x)^n = (nC0) + (nC1) x +(nC2) x^2 + (nC3)x^3 +...(nCn)x^n positioned x = -a million in the two sides (a million-a million)^n = (nC0)-(nC1)+(nC2)-(nC3)+...+ (nCn)(-a million)^n so (nC0)-(nC1)+(nC2)-(nC3)+...+(nCn)(-a million)^n = 0 (q.e.d.) Please one question at a time

Answer 5

(1+x)^n= C(n,0)+C(n,1)x+C(n,2)x^2+.............C(n,k)x^k+......................C(n,n)x^n
put x=1, we get the reqd. proof.