The value of (112 + 122 + 132 + 142 + ............. + 202) is:
- 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.
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