-  112 + 122 + 132 + ........... + 202
= (12 + 22 + 32 + ........... + 202) - (12 + 22 + 32 + ........... + 102)
= 20 (20 + 1) (40 + 1) 6 - 10 (10 + 1) (20 + 1) 6  = 2485.

The given expression is of the form (n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + ........ + (n+m)^2) where n= 112 and m= 90.

This is a particular type of series known as the Sum of Squares of natural numbers from n to (n+m). The formula to calculate such a series is given by:

Sum of squares of natural numbers from n to (n+m) = [(n+m)(n+m+1)(2(n+m)+1)]/6 - [(n)(n+1)(2n+1)]/6

Here, n = 112 and m = 90

Therefore, Sum of squares of natural numbers from 112 to 202 = [(202)(203)(405)]/6 - [(112)(113)(225)]/6

Sum of squares of natural numbers from 112 to 202 = [406406]/6 - [25200]/6

Sum of squares of natural numbers from 112 to 202 = 2485

Hence, the value of (112 + 122 + 132 + 142 + ............. + 202) is 2485.

If you think the solution is wrong then please provide your own solution below in the comments section .