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The proof can be obtained by the principle of mathematical induction method.

let the given statement be P(n):

(a+b)^n=nCoa^n+nC1a^n-1b+nC2a^n-2b^2+..................+nCna.b^n-1+nCnb^n

for n=1

P(1) : (a+b)^1 = 1Coa^1+1C1b^1=a+b

Therefore P(1) is true.

Let P(k) is true for some positive integer k,

(a+b)^k = kCoa^k+kC1a^k-1b+kC2a^k-2b^2+................+kCkb^k

Let P(k+1) is also true.

(a+b)^k+1 = k+1Coa^k+1 + k+1C1a^kb+k+1C2a^k-1b^2+............+k+1Ck+1b^k+1

now;

(a+b)^k+1 = (a+b)(a+b)^K

=(a+b)(kCoa^k+kC1a^k-1b+kC2a^k-2b^2 + ,..............+kCk

=kCoa^k+1+kC1a^kb+kC2a^k-1b^2+...........+kCk-1a^2b^k-1+kCkab^k+kCoa^kb+kC1a^k-1b^2+...............+kCk-1ab^k+kCkb^k+1

=kCoa^k+1+(kC1+kCo)a^kb+(kC2+kC1)a^k-1b^2+.........+(kCk+kCk-1)ab^k+kCkb^k+1

=k+1Coa^k+1+k+1C1a^kb+k+1C2a^k-1b^2+..............+k+1Ckab^k+k+1Ck+1b^k+1

therefore P(k+1) is also true

hence prooved

Hope it helps!!!

Thank you!!!