The Binomial Theorem says that: For all real numbers a and b and non-negative integers n, (a + b)... more The Binomial Theorem says that: For all real numbers a and b and non-negative integers n, (a + b) n = n r=0 n r a r b n−r. For example, (a + b) 0 = 1, (a + b) 1 = a + b, (a + b) 2 = a 2 + 2ab + b 2 , (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3. Proof. Let P (n) be the statement that for all real numbers a and b, (a + b) n = n r=0 n r a r b n−r. The Base Case is easy to establish. Now we prove the Inductive Step. Suppose that k ∈ Z ≥ is such that 'inductive hypothesis' (the formula for n = k, i.e., the statement P (k)) (a + b) k = k r=0 k r a r b k−r. (1) We want to prove that 'inductive conclusion' (the formula for n = k + 1, i.e., the statement P (k + 1)) (a + b) k+1 = k+1 r=0 k + 1 r a r b k+1−r. (2) We compute that (a + b) k+1 = (a + b) k · (a + b) (3) = k r=0 k r a r b k−r · (a + b) by the inductive hypothesis = k r=0 k r a r+1 b k−r + k r=0 k r a r b k+1−r by the distributive property; indeed, when we multiply a r b k−r in line 2 by a, the power of a increases by 1 to get a r+1 b k−r in the first term in line 3. Similarly, when we multiply a r b k−r by a, we get a r b k+1−r in the second term in line 3. Now a r b k+1−r in line 3 of (3) matches the form of the right-hand side of (2). To make the term a r+1 b k−r in line 3 of (3) also match, we shift the variable r down by 1 as follows. Define s = r + 1. Then r = s − 1. Moreover, when r is summed from 0 to k, we then have that s is summed from 1 to k + 1. So the first term in line 3 of (3) may be rewritten as k r=0 k r a r+1 b k−r = k+1 s=1 k s − 1 a s b k+1−s
The Binomial Theorem says that: For all real numbers a and b and non-negative integers n, (a + b)... more The Binomial Theorem says that: For all real numbers a and b and non-negative integers n, (a + b) n = n r=0 n r a r b n−r. For example, (a + b) 0 = 1, (a + b) 1 = a + b, (a + b) 2 = a 2 + 2ab + b 2 , (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3. Proof. Let P (n) be the statement that for all real numbers a and b, (a + b) n = n r=0 n r a r b n−r. The Base Case is easy to establish. Now we prove the Inductive Step. Suppose that k ∈ Z ≥ is such that 'inductive hypothesis' (the formula for n = k, i.e., the statement P (k)) (a + b) k = k r=0 k r a r b k−r. (1) We want to prove that 'inductive conclusion' (the formula for n = k + 1, i.e., the statement P (k + 1)) (a + b) k+1 = k+1 r=0 k + 1 r a r b k+1−r. (2) We compute that (a + b) k+1 = (a + b) k · (a + b) (3) = k r=0 k r a r b k−r · (a + b) by the inductive hypothesis = k r=0 k r a r+1 b k−r + k r=0 k r a r b k+1−r by the distributive property; indeed, when we multiply a r b k−r in line 2 by a, the power of a increases by 1 to get a r+1 b k−r in the first term in line 3. Similarly, when we multiply a r b k−r by a, we get a r b k+1−r in the second term in line 3. Now a r b k+1−r in line 3 of (3) matches the form of the right-hand side of (2). To make the term a r+1 b k−r in line 3 of (3) also match, we shift the variable r down by 1 as follows. Define s = r + 1. Then r = s − 1. Moreover, when r is summed from 0 to k, we then have that s is summed from 1 to k + 1. So the first term in line 3 of (3) may be rewritten as k r=0 k r a r+1 b k−r = k+1 s=1 k s − 1 a s b k+1−s
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Papers by victor effah