if `T_(N) ` denotes the nth term of the series 2+3+6+11+18+***, then `t_(50)` is

Options

A. `49^(2)-1`
B. `49^(2)`
C. `50^(2)+1`
D. `49^(2)+2`

Answer

D

Explanation

Let `T_(n)" be the "n^(th)` term of the given series. We observe that the successive differences of the terms form an A.P. So, let its `n^(th)` term be given by
`T_(n)=an^(2)+bn+c`
Putting n=1,2,3, we get
a+b+c=2,4a+2b+c=3 and, 9a+3b+c=6
Solving these equations, we get
a=1, b=-2, c=3
`:." "T_(n)=n^(2)-2n+3=(n-1)^(2)+2`
`rArr" "T_(50)=49^(2)+2`