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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 value of `t_(50)` is:

A

(a) `1075`

B

(b) `1175`

C

(c) `1275`

D

(d) `1375`

Text Solution

Verified by Experts

The correct Answer is:
C

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_(50)=(50xx51)/(2)=25xx51=1275`.
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Knowledge Check

  • Let n be an odd natural number of greater than 1. then the number of zeros at the end of the sum 999^(n)+1 is:

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