**Solution:** we have given a series , as : 2 + 3 + 6 + 11 + 18 + ...

Now, This difference of the terms of this series is in A.P.

3 - 2 = 1

6 - 3 = 3

11 - 6 = 5

18 - 11 = 7

So, the series obtained from the difference = 1,3,5,7,...

and to get back the original series we need to add the difference back to 2.

2+1 = 3,

2+1+3 = 6,

2+1+3+5= 11,

2+1+3+5+7 = 18 and so on.

So, we can say that n^{th} term of our given series ( 2 + 3 + 6 + 11 + 18+.... ) is = Sum of ( n - 1 ) term of series ( 1,3,5,7,... ) + 2

So, we need to calculate the sum of 49^{ }terms of the series 1,3,5,7,9,11,..

As we know formula for n^{th} term in A.P.

S_{n} = *n/*2[ 2a + ( n - 1 ) d ]

Here a = first term = 1 , n = number of term = 49 and d = common difference = 2 , So

S_{n} = 49/2[ 2( 1 ) + ( 49 - 1 ) 2 ] = 49 [ 1 + ( 49 - 1 ) ] = 49^{2}

Hence, Sum of 49 terms of series 1,3,5,7,9,11,.. = 49^{2}

Now, to get the T_{50} term.. add 2+ sum of the 1+3+5+7+..+97

So ,

**T**_{50} of series 2 + 3 + 6 + 11 + 18+....... = 2 + 49^{2} = 2 + 2401 = 2403