eeeeeeeKKK P4 proof by induction Watch
use the method of mathematical induction to prove the result given
n (sigma) r=1 r = 1/2 n(n+1)
(sorry that may not be the easiest form to read...maybe i'll paint it in word next time)
2un = 1/n - 1/n+1
2Sum[ui](i: 1->n) = (1/1 - 1/2) + (1/2 - 1/3) + .... + (1/n - 1/n+1)
2S = 1/1 - 1/n+1
S = n/2(n+1)
Let n=k+1, then:
So it's true for n=k+1 if it's true for n=k.
Now let n=1:
1 = (1/2)(1)(2) = 1
So it's true for n=1, and thus, by induction, true for all positive n.
ok guys have painted it in paint.
Induction method is:
Assume the theory is true for n = k. Then prove it's right for n = k+1 or n = k-1
When n = 1: Sum (n terms) = 1
By using the formula: Sum (n terms) = 1/2(1 + 1) = 2/2 = 1 ---> True.
Hence the formula is valid for the basis case where n = 1.
Step 2: Test the inductive step.
Assuming the formula to be true for all +ve integer n:
Sum (n + 1 terms) = Sum (n terms) + (n + 1)th term
= (n/2)(n + 1) + (n + 1)
= (n + 1)/2[n + 2]
= [(n + 1)/2][(n + 1) + 1]
Hence the formula is valid in finding the sum to (n + 1) terms. i.e.) (n + 1) has effectively been substituted for n in the formula.
Hence as the formula is proven valid for the basis case and the inductive step ---> By mathematical induction the formula is valid for all +ve integer n.