+0

# I need a hint on how to compute this:

+1
179
6

\((101+97)(101-97)+(93+89)(93-89)+\cdots +(5+1)(5-1)\)

Jun 25, 2020

#1
0

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!

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.

Jun 26, 2020
#4
0

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

∑[4*(198 - 16n), n, 0, 12 ] = 5,304

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

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)  =  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

Jun 26, 2020