# triangular numbers in Pascal's triangle triangular numbers = total numer of dots required to form a triangle Triangle 1 dot 3 dot 6 dot Triangular number 1 3 6 Show algebraically that the...

**triangular numbers in Pascal's triangle**

triangular numbers = total numer of dots required to form a triangle

Triangle

1 dot

3 dot

6 dot

Triangular number

1

3

6

Show algebraically that the nth triangular number is given by the formula

tn=n(n+1)/2

tn=triangle number

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The triangular numbers are : 1, 3, 6, .....n

1st term = 1 = 1

2nd term = 3 = 1 + 2

3rd term = 6 = 1 + 2 + 3

So, nth term `t(n) = 1 + 2 + 3 + ...(n-2) + (n-1) + n` ........(i)

Reversing all the terms of R.H.S. in (i) or writing them in descending order we get:

`t(n) = n + (n-1) + (n-2) + ... + 3 + 2 + 1` ..............(ii)

Adding (i) and (ii) we get:

`t(n) + t(n) = (n+1) + (n-1+2) + (n-2+3) + ... + (n-2+3) + (n-1+2) + (n+1)`

`rArr 2*t(n) = (n+1) + (n+1) + (n+1) + ... + (n+1) + (n+1) + (n+1)`

There are n terms of (n+1). So,

`2t(n) = n*(n + 1)`

`rArr t(n) =( n(n + 1))/2`

Hence, the nth triangular number is given by the formula:

`t(n) = (n(n + 1))/2`