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# Can someone give me a start to finish explenation on this?

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(a) Compute the sum 101^2 - 97^2 + 93^2 - 89^2 + .... + 5^2 - 1^2
(b) Compute the sum (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + ... + (a+d)^2 - a^2

Aug 27, 2020

#1
+1038
+1

Didn't you just ask this like 3 minutes ago?

Also, it has been answered here. But, finding that you probably don't have the time to go over to the link, as you are too busy posting questions every 3 minutes, here is the answer guest has posted:

(a) 101^2 - 97^2 + 93^2 - 89^2 + ... + 5^2 - 1^2 = 101 + 97 + 93 + 89 + ... + 5 + 1 = 1326.

(b) (a + (2n + 1)d)^2 - (a + (2n)d)^2 + ... + (a + d)^2 - a^2 = (a + (2n + 1) d + (a + 2nd + ... + a + d) + a = n(3a + (n + 2)d).

:)

Aug 27, 2020
#2
0

sorry, my computer was lagging so I accidentally posted several, but can you give an explenation so I can do more of these?

Aug 27, 2020
#3
+1038
+1

I'm sorry, but I am also quite confused on how the guest got the answer. I did it with difference of squares, which is a long, rigorous process... let me sent this to someone else...

ilorty  Aug 27, 2020
#4
+1

a ) is wrong!

Here is your entire sequence grouped in "twos"

(10201 -9409, 8649 -7921, 7225 -6561, 5929 -5329, 4761 -4225, 3721 -3249, 2809 -2401, 2025 -1681, 1369 -1089, 841 -625, 441 -289, 169 -81, 25 -1)

What is the difference between each pair beginning with first? The following:

(792, 728, 664, 600, 536, 472, 408, 344, 280, 216, 152, 88, 24) - This is an arithmetic sequence.                    How do you sum it up?

[F + L] / 2 * N (number of terms) =Sum, where F=First term, L=Last term, N =Number of terms.

[792 + 24] / 2 * 13 =5,304 - And that is it!.

Aug 27, 2020
#5
+1038
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Im sorry. I was really skeptical about the other answer....but I didn't have a better way...so....

ilorty  Aug 27, 2020
#6
+1

b) See the answer in LaTex here:  https://web2.0rechner.de/mitglieder/heureka/?answerpage=531

Aug 27, 2020
#7
+111563
+1

Thanks guest, here is a little more insight for others to see what you have done.

(a) Compute the sum 101^2 - 97^2 + 93^2 - 89^2 + .... + 5^2 - 1^2

I paired them first

(101^2 - 97^2) + (93^2 - 89^2) + .... + (5^2 - 1^2)

then turned them around

(5^2 - 1^2)+ ..................   +  (93^2 - 89^2) +  (101^2 - 97^2)

$$T_1=(5^2 - 1^2) = [(3+2)^2 - (3-2)^2]\\ T_2=(13^2-9^2)= [(11+2)^2 - (11-2)^2]\\ ...\\ T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\ ...\\ T_{13}=(101^2-97^2)\\\quad= [(8*13-5+2)^2 - (8*13-5-2)^2]\\ \quad= [(99+2)^2 - (99-2)^2]\quad \text{just checking... it is good}\\$$

$$T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\ T_n= [(8n-3)^2 - (8n-7)^2]\\ T_n= [(8n)^2-48n+9)] - [(8n)^2-112n+49)]\\ T_n= [-48n+9)] - [-112n+49)]\\ T_n= -48n+9 +112n-49\\ T_n= 64n-40\\ \text{This is an AP}\\ a=24, d=64,\;n=13\\ S_n=\frac{n}{2}(2a+(n-1)d]\\ S_{13}=\frac{13}{2}(2*24+12*64]\\ S_{13}=13(24+6*64]\\ S_{13}=13*408\\ S_{13}=5304\\$$

LaTex:

T_1=(5^2 - 1^2) = [(3+2)^2 - (3-2)^2]\\
T_2=(13^2-9^2)=  [(11+2)^2 - (11-2)^2]\\
...\\
T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\
...\\
T_{13}=(101^2-97^2)\\\quad=  [(8*13-5+2)^2 - (8*13-5-2)^2]\\
\quad=  [(99+2)^2 - (99-2)^2]\quad \text{just checking... it is good}\\

T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\
T_n= [(8n-3)^2 - (8n-7)^2]\\
T_n= [(8n)^2-48n+9)] - [(8n)^2-112n+49)]\\
T_n= [-48n+9)] - [-112n+49)]\\
T_n= -48n+9 +112n-49\\
T_n= 64n-40\\
\text{This is an AP}\\
a=24, d=64,\;n=13\\
S_n=\frac{n}{2}(2a+(n-1)d]\\
S_{13}=\frac{13}{2}(2*24+12*64]\\
S_{13}=13(24+6*64]\\
S_{13}=13*408\\
S_{13}=5304\\

Aug 27, 2020
#8
+93
+2

Wow... I really need to work on reading LaTeX!

joliel3  Aug 27, 2020
#9
+111563
0

Why ?

You do not need to understand latex.

The latex coding that i have posted is for my benefit, and for the possible benefit of someone who specifically wants to learn LaTex, it is not a part of the answer.

The answer is written in normal maths symbols.

(it is rendered LaTex, but it is not LaTex coding)

Melody  Aug 27, 2020