# eeeeeeeKKK P4 proof by induction Watch

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i am helpless at this!!!!! some one help! please! lol

use the method of mathematical induction to prove the result given

n (sigma) r=1 r = 1/2 n(n+1)

thank you.

(sorry that may not be the easiest form to read...maybe i'll paint it in word next time)

use the method of mathematical induction to prove the result given

n (sigma) r=1 r = 1/2 n(n+1)

thank you.

(sorry that may not be the easiest form to read...maybe i'll paint it in word next time)

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#2

un = 1/2n(n+1) = 1/2 (1/n - 1/n+1)

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)

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)

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#3

Assume it's true for n=k, then:

1+2+3+...+k=(1/2)k(k+1)

Let n=k+1, then:

1+2+3+...+k+(k+1)=(1+2+3+...+k)+ (k+1)=(1/2)k(k+1)+(k+1)=(k+1)((k/2)+1)=(k+1)(k+2)/2=(1/2)(k+1)(k+2)

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.

1+2+3+...+k=(1/2)k(k+1)

Let n=k+1, then:

1+2+3+...+k+(k+1)=(1+2+3+...+k)+ (k+1)=(1/2)k(k+1)+(k+1)=(k+1)((k/2)+1)=(k+1)(k+2)/2=(1/2)(k+1)(k+2)

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.

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#8

(Original post by

ok guys have painted it in paint.

please help!!!!

**posh_git**)ok guys have painted it in paint.

please help!!!!

Induction method is:

Assume the theory is true for n = k. Then prove it's right for n = k+1 or n = k-1

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lol sorry obviously i dind't read. lol will add u 2 the list of "got to give rep to" people - both dvs and bchl85!!!!!!!

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#10

Step 1: Test the basis case.

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.

Nima

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.

Nima

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