Question

If $t_n$ denotes the $n$th term of the series $2+3+6+11+18+...$then $t_{50}$ is

• A. ${49}^2-1$
• B. ${49}^2$
• C. ${50}^2+1$
• D. ${49}^2+2$

Answer 1

Answer: (D)

${49}^2+2$

In the given series, the difference between the successive terms are in A.P.

Therefore, $t_n$ will be of the form $a{n}^2+bn+c$

Now,

$t_1=2=a+b+c$

$t_2=3=4a+2b+c$

$t_3=6=9a+3b+c$

Solving for $a$, $b$ and $c$

We get $a=1$, $b=-2$ and $c=3$

Therefore, $t_n=n^2-2n+3=(n-1{)}^2+2$

Thus, $t_{50}={49}^2+2$
Ans: D