SOLUTION: Use mathematical induction to prove that: 1+2+3+...+n=[n(n+1)]/2

Question 1053689: Use mathematical induction to prove that:
1+2+3+...+n=[n(n+1)]/2
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Use mathematical induction to prove that:
1+2+3+...+n=[n(n+1)]/2
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Show it is true for n = 1::
1 = [1(1+1)]/2 = 2/2 = 1
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Assume it is true for n = k::
1+2+3+...+k = [k(k+1)]/2
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Prove it is true for n = k+1::
1 + 2 + 3 +...+ k + (k+1) = [k(k+1)]/2 + (k+1)
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= [(k^2+k]/2 + (k+1)
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= (k^2 + k + 2k + 2)/2
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= (k^2 + 3k + 2)/2
------
= [(k+1)(k+2)]/2
------
= [(k+1)((k+1)+1)]/2
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QED
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Cheers,
Stan H.
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Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
See the lesson
    - Mathematical induction and arithmetic progressions
in this site.

Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Method of Mathematical induction".

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