Alternative Proof of Binomial Theorem for Positive Integral Index (Combinatorial Method)
Alternative Proof of Binomial Theorem for Positive Integral Index (Combinatorial Method). We have, (a + b) n = (a + b) (a + b) ....... n times. The terms on the RHS are obtained by taking one letter from each factor and multiplying them together. Choosing 'a' fr..
Proof:
Since |x|<1, we have by binomial theorem, Comparing, the coefficients of y in (1) and (2), we get ..
Since |x|<1, we have by binomial theorem, Comparing, the coefficients of y in (1) and (2), we get ..Binomial Theorem for Fractional Index
For any rational number n, We accept this expansion without proof..
For any rational number n, We accept this expansion without proof..Binomial Theorem
1. A sentence is called a statement if it can be adjudged as true or false. Every statement is a sentence, but a sentence may or may not be a statement. 2. A statement involving natural number n is generally denoted by P(n). 3. A binomial is an algebraic expression o..
Proof:
If P coincides with O then AO = OB or AP = PB. . If P does not coincides O, then compare triangles AOP and BOP. PA=PB ( given) AO=OB ( O is the mid point of AB) OP=OP ( common side) But ( Linear pair) ..
If P coincides with O then AO = OB or AP = PB. . If P does not coincides O, then compare triangles AOP and BOP. PA=PB ( given) AO=OB ( O is the mid point of AB) OP=OP ( common side) But ( Linear pair) ..Proof:
If Q coincides with O, then AO = BO. i.e., AQ=BQ If Q is distinct i.e., Q does not coincides with O, then compare triangles AOQ and BOQ. AO=BO ( O is the mid point of AB) (given) OQ=OQ ( common side) AQ=BQ ( CP..
If Q coincides with O, then AO = BO. i.e., AQ=BQ If Q is distinct i.e., Q does not coincides with O, then compare triangles AOQ and BOQ. AO=BO ( O is the mid point of AB) (given) OQ=OQ ( common side) AQ=BQ ( CP..Proof:
x + 1 = 1 (Theorem 2a) In particular for x = 0, we have 0 + 1 = 1 (1) x.0 = 0 (Theorem 2b) In particular, for x =1 1.0 = 0 0.1 = 0 (3a) (2) From (1) and (2), we have For 0 B, 0 + 1 = 1 0.1 = 0 1 is the complement of 0. 0' = 1 ..
x + 1 = 1 (Theorem 2a) In particular for x = 0, we have 0 + 1 = 1 (1) x.0 = 0 (Theorem 2b) In particular, for x =1 1.0 = 0 0.1 = 0 (3a) (2) From (1) and (2), we have For 0 B, 0 + 1 = 1 0.1 = 0 1 is the complement of 0. 0' = 1 ..Proof:
Let n > 0. When n = 0, tan (0 p + x) = tan x which is true. The theorem is proved by mathematical induction when n=1, tan ( p +x)=tan p =RHS. Let the theorem be true for n = m > 0 then tan(m p + x) = tan p Since the theorem is true for all n = 1, n = m,..
Let n > 0. When n = 0, tan (0 p + x) = tan x which is true. The theorem is proved by mathematical induction when n=1, tan ( p +x)=tan p =RHS. Let the theorem be true for n = m > 0 then tan(m p + x) = tan p Since the theorem is true for all n = 1, n = m,..Proof:
x + 1 = 1 (Theorem 2a) In particular, for x = 0 0 + 1 =1 1 + 0 = 1 (1) (Axiom 3a) x . 0 = 0 In particular for 1 B, we have 1 . 0 = 0 (2) From (1) and (2), we have 0 is the complement of 1. (Axiom ..
x + 1 = 1 (Theorem 2a) In particular, for x = 0 0 + 1 =1 1 + 0 = 1 (1) (Axiom 3a) x . 0 = 0 In particular for 1 B, we have 1 . 0 = 0 (2) From (1) and (2), we have 0 is the complement of 1. (Axiom ..Proof:
From the above theorem, we have = 1 \ Using Sandwich theorem, we ge..
From the above theorem, we have = 1 \ Using Sandwich theorem, we ge..See what our Users say :
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