Question

Mathematical induction to show the sum of natural numbers n

Answer 1

Proof (Pictoral):



Represent each natural number n by n-many unit squares. Then each natural number is equal to the area of its unit squares. For each n we can construct a "pyramid" that is n squares high in the manner as depicted above. Then the area of the nthpyramid is the sum 1+2+...+n.

But the nth pyramid is contained in the n(n+1) rectangle and clearly its area is half the rectangle, so:

(1)

1+2+...+n=n(n+1)2■

Proof (Algebraic): We prove this by mathematical induction. Let P(n) be the statement that the sum of the first n natural numbers is equal to n(n+1)2. If n=1then P(1) says that the sum of the first 1 natural numbers is equal to 1(2)2=1 which is true.

Assume that for some k that P(k) is true. We want to show that P(k+1) is true. We have that the sum of the first k+1 number is:

(2)

1+2+...+k+(k+1)=[1+2+...+k]+(k+1)

By assumption P(k) is true and so 1+2+...+k=k(k+1)2. So:

(3)

1+2+...+k+(k+1)