Question For n 1 derive each of the identities below an 0

Question: For n > = 1, derive each of the identities below : (a)(n 0) +(n 1) +(n 2) +...+ (n n) =2^n. [Hint: Let a = b = 1 in the binomial theorem.] (b) (n 0) -(n 1) + (n 2)- ?+(-1)^n(n n) = 0.

Solution

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

oR

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.

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