Question

Show that no triangular number can be of the form (3n - 1).



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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

The first triangular number consist of 1 .

Then we write 3 , then 6 and then 10 and so on .

First triangular number is 1 and then we add 2 , 1 + 2 = 3 , and then we add 3 to get 3 + 3 = 6 and then 4 to get 6 + 4 = 10 and so on .

Let S be a triangular number .

S = 1 + 2 + 3 ....... n

⇒ S = 1 + 2 + ... n - 1 + n

⇒ S = n + 1 + n + 1 + ,,,,,, n/2 times

⇒ S = n/2 ( n + 1 )

Now let us assume that a triangular number is of the form 3 n - 1 .

So we get :

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

⇒ 3 n - 1 = n²/2 + n/2

⇒ 3 n - n/2 = n²/2 + 1

⇒ ( 6 n - n ) / 2 = ( n² + 2 ) / 2

⇒ 5 n / 2 = ( n² + 2 ) / 2

⇒ 5 n = n² + 2

⇒ n² - 5 n + 2 = 0

Here √(b² - 4 a c ) ⇒ Δ is not an integer .

√[ (-5)² - 4×2×1 ]

⇒ √( 25 - 8 )

⇒ √17

Hence there is a contradiction ,

So the triangular numbers cannot be of the form 3 n - 1 .

Step-by-step explanation:

This can only be proved by the contradiction method .

We assume the triangular numbers to be of the form 3 n - 1 .

We already know that the triangular numbers are of the form n/2 ( n + 1 ) which is proved above .

Hence we can prove by equating both and when there will be a contradiction , our proof ends .