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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`.
The number of positive integers lying between `t_(100)` and `t_(101)` are:

A

(a) `99`

B

(b) `100`

C

(c) `101`

D

(d) `102`

Text Solution

Verified by Experts

The correct Answer is:
B
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)`
`t_(100)=(100xx101)/(2)=5050`
`t_(101)=(101xx102)/(2)=101xx51=5151`
Number of positive integers lying between `t_(100)` and `t_(101)`
`=5151-5050-1`
`=101-1=100`.
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