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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

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