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The numbers 1,3,6,10,15,21,28"..." are c...

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

A

(a) `1+sqrt((m^(2)+2m))`

B

(b) `1+sqrt((m^(2)+2))`

C

(c) `1+sqrt((m^(2)+m))`

D

(d) None of these

Text Solution

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The correct Answer is:
D
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)`.
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