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  1. Prove by mathematical induction that for each positive integer n, the sum of the first n triangular numbers is given by the formula: click on link for formular. http://i57.tinypic.com/zoky0.jpg

 Feb 24, 2015

Best Answer 

 #1
avatar+130516 
+5

We wish to show that....

1 + 2 + 3 + 4 +.....+ (n-1) + n =  n(n + 1) /2

Show it's true for n= 1

(1)(1+1)/2= 2/2  = 1

Assume it's true for k, that is    1 + 2 + 3 + 4 + ..... +( k - 1) + k  = k(k+1)/2

Prove it's true for k + 1

That is  .....1 + 2 + 3 + 4 +......+ k + (k+1)  = (k+1)(k+2)/2

We have

1 + 2 + 3 + 4 +......+ k + (k+1 ) = k(k+1)/2 + (k+1) =

1 + 2 + 3 + 4 +......+ k + ( k+1)  = k(k+1)/2 + 2(k+1)/2 =

1 + 2 + 3 + 4 +......+ k + (k+1)  =  [k(k+1) + 2(k+1)] / 2 =  {factor out (k + 1) }

1 + 2 + 3 + 4 +......+ k + (k+1)  = [(k +1)(k + 2)] / 2

And that's what we wished to prove.....!!!

 

 Feb 24, 2015
 #1
avatar+130516 
+5
Best Answer

We wish to show that....

1 + 2 + 3 + 4 +.....+ (n-1) + n =  n(n + 1) /2

Show it's true for n= 1

(1)(1+1)/2= 2/2  = 1

Assume it's true for k, that is    1 + 2 + 3 + 4 + ..... +( k - 1) + k  = k(k+1)/2

Prove it's true for k + 1

That is  .....1 + 2 + 3 + 4 +......+ k + (k+1)  = (k+1)(k+2)/2

We have

1 + 2 + 3 + 4 +......+ k + (k+1 ) = k(k+1)/2 + (k+1) =

1 + 2 + 3 + 4 +......+ k + ( k+1)  = k(k+1)/2 + 2(k+1)/2 =

1 + 2 + 3 + 4 +......+ k + (k+1)  =  [k(k+1) + 2(k+1)] / 2 =  {factor out (k + 1) }

1 + 2 + 3 + 4 +......+ k + (k+1)  = [(k +1)(k + 2)] / 2

And that's what we wished to prove.....!!!

 

CPhill Feb 24, 2015

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