If `t_n` denotes the nth term of the series `2+3+6+11+18+...`then `t_(50)`=......

Options

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

Answer

D

Explanation

Let `S_(n)` be sum of the series 2+3+6+11+18+***+`t_(50).`
`therefore s_(n) =2+3+6+11+18***+t_(50)`
`and S_(n) =0+2+3+6+11+18***+t_(49) +y_(50)`
On subtracting Eq (ii) from eq (i) we get
`0=2+1+3+5+7+***` upto 49 terms
`t_(50) =2+[1+3+55+7+***`upto 49 terms]
`=2+(49)/(2)[2xx1++48xx2]`
`=2+(49)/(2)xx[2+96]`
`=2+[49+49xx48]`
`2+49xx49=2+(49)^(2)`