Question

If $T_n$ denotes the $nth$ term of the series $2 + 3 + 6 + 11 + 18 + ...... ,$ then $T_{50}$ is equal to

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

Answer 1

Answer: (A)

$49^2+2$

Answer 2

Let $S_{n}= 2 + 3 + 6 +11+ 18 +......+ T_{n}$ Also $S_{n}= 2 + 3 + 6 + 11+......+ T_{n-1} + T_{n}$ $\therefore 0 = 2 + \left(1 + 3 + 5 + 7 + .... {\text{ upto}} \left(n - 1\right){\text{ term}}\right) - T_{n}$ $\therefore T_{n} = 2+ 1+ 3 + 5 + 7 + ...... {\text{upto}} \left(n - 1\right) {\text{terms}}$ $= 2+\frac{n-1}{2}\left[2.1+\left(n-2\right)2\right]$ $=2+\left(n-1\right)\left[1+n-2\right]$ $= 2+\left(n-1\right)\left(n-1\right)$ $\therefore T_{50} = 49^{2}+2$