#1**+2 **

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!

CalTheGreat Jun 25, 2020

#2**-1 **

Welcome back Cal! I think the numbers increase by 64 each time...? I don't know how that's supposed to help me tho..

AnimalMaster
Jun 25, 2020

#3**+1 **

Look for a pattern in the (x+y) terms. The (x-y) terms remain the same.

gwenspooner85 Jun 26, 2020

#4**0 **

Use this "closed form" formula to sum them up;

∑[**4*(198 - 16n)**, n, 0, 12 ] = 5,304

Guest Jun 26, 2020

#5**+1 **

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.

raspberrypitechguy Jun 26, 2020

#6**0 **

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) = 101^{2} - 97^{2} = 792

(93 + 89)(93 - 89) = 93^{2} - 89^{2 } = 729

...

(5 + 1)(5 - 1) = 5^{2} - 1^{2} = 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}

= 64n^{2} + 80n + 25) - (64n^{2} + 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**

geno3141 Jun 26, 2020