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

I would prefer it if the answers are not in latex. Thank you so much for helping me

Dec 3, 2019

#1
0 Question in LaTeX format. welcom

Dec 3, 2019
#2
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wait why does it say waiting for moderation?

Guest Dec 3, 2019
#3
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never mind i got it

do you know the answer though? it's ok even if you can answer one

Guest Dec 3, 2019
#4
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for a)

I THINK you use the special case a2-b2 = (a-b)(a+b)

Guest Dec 3, 2019
#5
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a)     sum_(n=1)^26 (-1)^(n + 1) (105 - 4 n)^2 = 5304

Dec 3, 2019
#6
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I don't get it is 5304 the answer for the first one? and could you explain a bit more on how you got there?

Dec 3, 2019
#7
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(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).

Dec 4, 2019