Proof by induction - Factorials
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Proof by Induction Prove that: n ∑ 1 (r 2 + 1)r!= n(n + 1)! Case for n = 1: 2 × 1=1 × 2 It works! Which is nice. Assuming this beast is true when n = k: k ∑ 1 (r 2 + 1)r!= k(k + 1)! Now working out what’s going on when n = k + 1: (Note that we know, by the definition of what sums are, that, p+1 ∑ 1 f (r)= p ∑ 1 f (r)+ f (p + 1) (Also note that we know that, by the definition of what a factorial is: (q + 1)! = (q + 1)(q!) ∴ k+1 ∑ 1 (r 2 + 1)r! = k ∑ 1 (r 2 + 1)r! + ((k + 1) 2 + 1)(k + 1)! = k(k + 1)! + (k 2 +2k + 2)(k + 1)! = (k 2 +3k + 2)(k + 1)! = (k + 2)(k + 1)(k + 1)! = (k + 1)(k + 2)! ∴ by induction, ∑ n 1 (r 2 + 1)r!= n(n + 1)!. 1
description
A worked-example of an A-level standard maths question on proof by induction, involving an expression containing factorials.
Transcript of Proof by induction - Factorials
Proof by Induction
Prove that:
n∑1
(r2 + 1)r! = n(n+ 1)!
Case for n = 1:
2× 1 = 1× 2
It works! Which is nice.Assuming this beast is true when n = k:
k∑1
(r2 + 1)r! = k(k + 1)!
Now working out what’s going on when n = k + 1:(Note that we know, by the definition of what sums are, that,
p+1∑1
f(r) =
p∑1
f(r) + f(p+ 1)
(Also note that we know that, by the definition of what a factorial is:
(q + 1)! = (q + 1)(q!)
∴k+1∑1
(r2 + 1)r! =k∑1
(r2 + 1)r! + ((k + 1)2 + 1)(k + 1)!
= k(k + 1)! + (k2 + 2k + 2)(k + 1)!
= (k2 + 3k + 2)(k + 1)!
= (k + 2)(k + 1)(k + 1)!
= (k + 1)(k + 2)!
∴ by induction,∑n
1 (r2 + 1)r! = n(n+ 1)!.
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