The numbers `1,3,6,10,15,21,28."……"` are called triangular numbers. Let `t_(n)` denotes the bth triangular number such that `t_(n)=t_(n-1)+n,AA n ge 2`.
If `(m+1)` is the nth triangular number, then `(n-m)` is

Options

A. `1+sqrt((m^(2)+2m))`
B. `1+sqrt((m^(2)+2))`
C. `1+sqrt((m^(2)+m))`
D. None of these

Answer

D

Explanation

Given sequece `1,3,6,10,15,21,28,"…"`
where `t_(n)=t_(n-1)+n,AA n ge 2`
So, `t_(n)=[t_(n-2)+(n-1)]+n`
`vdots " "vdots " " vdots " "`
`t_(n)=t_(1)+2+3+"........"+(n-1)+n`
`t_(n)=1+2+3+"......"+n`
`t_(n)=(n(n+1))/(2)`
According to the question, `(m+1)` is the nth triangular number, then
`(n(n+1))/(2)=m+1`
`n^(2)+n-2(m+1)=0`
`n=(-1pmsqrt(1+8(m+1)))/(2)`
`=(-1+sqrt((8m+9)))/(2)`
`n-m=(-1+sqrt(8m+9-2mn))/(2)`.