Here's a hint!
(101+97)(101-97)=792
(93+89)(93-89)=728
(5+1)(5-1)=24
Notice that the end numbers are always 4, and the pattern. (How the first number decreases)
Hope this helps!
Welcome back Cal! I think the numbers increase by 64 each time...? I don't know how that's supposed to help me tho..
Look for a pattern in the (x+y) terms. The (x-y) terms remain the same.
Use this "closed form" formula to sum them up;
∑[4*(198 - 16n), n, 0, 12 ] = 5,304
101 + 97 = 198
93 + 89 = 182
Subtract 198 from 182, so 16 is the common length.
so we have 4(198 + 182 + ... + 6)
This is an arithmatic sequence so
(6 + 198)/2 * number in sequence
We know we add 12 each time so 198-6 = 192, 192/12 = 16
+ 1 for the first term
204/2 , * 17
Then, multiply 4, and that is your answer.
If you look at the "first terms", the 101, 93, ... 5, these have a difference of 8.
We can use this difference to express not only the "first terms" (101, 93, ... 5) but also the "second terms"
(97, 89, ... 1):
If n = 0: 8n + 5 = 5 8n + 1 = 1
...
If n = 11, 8n + 5 = 93 8n + 1 = 89
If n = 12, 8n + 5 = 101 8n + 1 = 97
Also note that if you multiply the expressions together, you get a difference of squares:
(101 + 97)(101 - 97) = 1012 - 972 = 792
(93 + 89)(93 - 89) = 932 - 892 = 729
...
(5 + 1)(5 - 1) = 52 - 12 = 24
If we rewrite the numbers using the 8n + 5 and 8n + 1 forms;
[ (8n + 5) + (8n + 1) ] ·[ (8n + 5) - (8n + 1) ]
= (8n + 5)2 - (8n + 1)2
= 64n2 + 80n + 25) - (64n2 + 16n + 1)
= 64n + 24 [this clearly indicates that it will be an arithmetic progression with a common difference of 64]
First term: n = 0 ---> value = 24
Last term: n = 12 ---> value = 792 [there are 13 term]
Sum = (13)(24 + 792) / 2 = 5304