+306
Compute the sum 1012−972+932−892+⋯+52−12 by pairing the terms and applying the difference of squares.
+23254 1012 - 972 = (101 - 97)(101 + 97) = 4·198
932 - 892 = (93 - 89)(93 + 89) = 4·182
852 - 812 = (85 - 81()(85 + 81) = 4·166
...
52 - 12 = (5 -1)(5 + 1) = 4·6
Sum = 4·198 + 4·182 + 4·166 + ... + 4·6 = 4·[ 198 + 182 + 166 + ... + 6 ]
198 + 182 + 166 + ... + 6 is an arithmetic series whose first term = 198, last term = 6, common difference = -16
The formula for the sum is: Sum = n · (t1 + tn) / 2
We'll need to find the number of terms; we can do this by using this formula: tn = t1 + (n - 1)·d
---> 6 = 198 + (n - 1)·-16 ---> -192 = (n - 1)·-16 ---> 12 = n - 1 ---> n = 13
Sum = n · (t1 + tn) / 2 ---> Sum = 13 · (198 + 6) / 2 ---> 1326
So, 4·[ 198 + 182 + 166 + ... + 6 ] = 4·1326 = 5304
