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 nth 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 nth term in A.P.
Sn = n/2[ 2a + ( n - 1 ) d ]
Here a = first term = 1 , n = number of term = 49 and d = common difference = 2 , So
Sn = 49/2[ 2( 1 ) + ( 49 - 1 ) 2 ] = 49 [ 1 + ( 49 - 1 ) ] = 492
Hence, Sum of 49 terms of series 1,3,5,7,9,11,.. = 492
Now, to get the T50 term.. add 2+ sum of the 1+3+5+7+..+97
So ,
T50 of series 2 + 3 + 6 + 11 + 18+....... = 2 + 492 = 2 + 2401 = 2403